Course Theory
Book 1. Foundations of Olympiad Algebra
Book 1. Foundations of Olympiad Algebra
- 1. Algebraic Expressions and Identities
- 2. Factorisation Methods
- 3. Equations and Systems
- 4. Polynomials I
- 5. Sequences and Recurrences
- 6. Algebraic Transformations in Problems
- 7. Introductory Inequalities
- 8. Introductory Functional Equations
- 9. Algebraic Number Problems
- 10. Mixed Algebra Problems I
- 11. Strategy Notes
- 12. Mock Olympiads I
Chapter
Algebraic Expressions and Identities
Key Idea
Olympiad algebra does not begin with mechanical expansion. It begins with recognising the shape of an expression. A good transformation reduces chaos: it reveals a common factor, a difference of squares, a sum of cubes, symmetry, or a useful substitution.
In this module, an expression should be read as a structure: which parts repeat, what resembles a square, where a difference is hidden, whether \(x+y\) and \(xy\) can be used, and whether a condition like \(a+b+c=0\) is present.
Basic Facts
Basic identities: \[ a^2-b^2=(a-b)(a+b), \] \[ a^3-b^3=(a-b)(a^2+ab+b^2), \] \[ a^3+b^3=(a+b)(a^2-ab+b^2). \]
Also useful: \[ (a+b)^2=a^2+2ab+b^2,\quad (a-b)^2=a^2-2ab+b^2, \] \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \]
When to Use This Method
Use identities when you need to prove divisibility, factor an expression, show that a number is composite, solve a system through \(x+y\) and \(xy\), prove non-negativity, or replace a long expression by a short one.
How to Recognise the Method
A difference of two similar squares almost always asks for \(u^2-v^2\). Cubes with signs ask for sum or difference of cubes. A symmetric expression in two variables often depends only on \(s=x+y\) and \(p=xy\). The condition \(a+b+c=0\) almost always means \(a^3+b^3+c^3=3abc\).
Typical Mistakes
Do not expand everything automatically: full expansion often destroys structure. Do not confuse \(a^3+b^3\) with \((a+b)^3\). Do not forget to check cases where a common factor may be zero. In integer problems, after factoring, check that both factors are really greater than \(1\) when proving compositeness.
Mini-Checklist
1. Is there a difference of squares? 2. Is there a sum or difference of cubes? 3. Can terms be grouped in pairs? 4. Is the expression symmetric? 5. Does the substitution \(s=x+y\), \(p=xy\) help? 6. Can the goal be proved through a product of factors?
Example 1. Hidden Difference of Squares
First look for two squares; expand only if necessary.
Problem. Factor \( (x^2+3x+1)^2-(x^2+x+1)^2 \).
This is a difference of squares: \(U^2-V^2=(U-V)(U+V)\), where \(U=x^2+3x+1\), \(V=x^2+x+1\).
We get \(U-V=2x\), \(U+V=2x^2+4x+2=2(x+1)^2\). Hence the expression is \(4x(x+1)^2\).
Comment. Full expansion is longer and hides the structure.
Example 2. Cubes and Divisibility
Sum and difference of cubes are often used to prove divisibility.
Problem. Prove that if \(a-b\) divides \(a+b\), then \(a-b\) divides \(a^3-b^3\).
By the difference of cubes, \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Therefore \(a^3-b^3\) is divisible by \(a-b\) for all integers \(a,b\), provided \(a\ne b\).
The condition about \(a+b\) is unnecessary; the key is to notice the correct factor.
Comment. Olympiad problems sometimes contain extra information; do not force it into the solution.
Example 3. Grouping
If an expression does not match one formula, group terms so that common factors appear.
Problem. Factor \(x^3+x^2y-xy^2-y^3\).
Group the first two and last two terms: \[ x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y). \] Factor out \(x+y\):
\((x+y)(x^2-y^2)=(x+y)(x-y)(x+y)=(x-y)(x+y)^2\).
Comment. Grouping often creates a difference of squares.
Example 4. Symmetric Substitution
In two variables, symmetry often suggests \(s=x+y\), \(p=xy\).
Problem. Express \(x^2+y^2\) and \(x^3+y^3\) through \(s=x+y\) and \(p=xy\).
We have \(x^2+y^2=(x+y)^2-2xy=s^2-2p\).
Also \(x^3+y^3=(x+y)(x^2-xy+y^2)=s((x^2+y^2)-p)=s(s^2-3p)\).
Comment. This substitution is essential for symmetric systems and equations.
Example 5. The Condition \(a+b+c=0\)
This is one of the most useful conditions in olympiad algebra.
Problem. Prove that if \(a+b+c=0\), then \(a^3+b^3+c^3=3abc\).
Use the identity \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] If \(a+b+c=0\), the right side is zero.
Therefore \(a^3+b^3+c^3-3abc=0\), so \(a^3+b^3+c^3=3abc\).
Comment. This identity should be remembered as a tool, not as an isolated fact.
Example 6. Compositeness by Factoring
Factoring helps prove that an expression cannot be prime.
Problem. Prove that for every integer \(n>1\), the number \(n^4+4\) is composite.
Use Sophie Germain's trick: \[ n^4+4=n^4+4n^2+4-4n^2=(n^2+2)^2-(2n)^2. \] Hence \[ n^4+4=(n^2-2n+2)(n^2+2n+2). \]
For \(n>1\), both factors are greater than \(1\), so the number is composite.
Comment. Do not forget to check the sizes of the factors.
Example 7. Non-Negativity as a Sum of Squares
Even in an identities module, it is useful to recognise sums of squares.
Problem. Prove that \(a^2+b^2+c^2-ab-bc-ca\ge 0\) for all real \(a,b,c\).
Multiply the expression by \(2\): \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] The right side is non-negative.
Hence the original expression is non-negative. Equality holds only when \(a=b=c\).
Comment. This form often appears next to \(a^3+b^3+c^3-3abc\).
Example 8. A System via Sum and Product
A symmetric system should not be solved directly.
Problem. Find all real \(x,y\) such that \(x+y=5\), \(x^2+y^2=13\).
From \(x^2+y^2=(x+y)^2-2xy\), we get \(13=25-2xy\), hence \(xy=6\).
Now \(x\) and \(y\) are roots of \(t^2-5t+6=0\). Thus \(t=2\) or \(t=3\). Therefore \((x,y)=(2,3)\) or \((3,2)\).
Comment. This is the first step toward olympiad-style symmetric systems.
Chapter
Factorisation Methods
Key Idea
Factorisation in olympiad algebra is not only a way to simplify an expression. It is a way to reveal hidden reasons for divisibility, compositeness, equality to zero, or impossibility.
The main question of the module is: which factor should appear, and why?
Basic Facts
The most common tools are a common factor, grouping, quadratic factorisation, difference of squares, sum and difference of cubes, and the identity
\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).\]
For polynomials, remember: if \(P(r)=0\), then \(x-r\) is a factor of \(P(x)\).
When to Use This Method
Factorisation is useful when the task asks for divisibility, integer solutions, compositeness, equations involving products, or a non-negative factor.
In parameter problems, first look for the parameter value that creates the required root or common factor.
How to Recognise the Method
Look for repeated blocks, forms \(U^2-V^2\), homogeneity, symmetry, vanishing at \(a=b\), \(b=c\), or \(x=r\). If an expression becomes zero under a simple substitution, it often has a hidden linear factor.
Typical Mistakes
Do not expand everything immediately: expansion may hide the structure. When proving compositeness, do not forget to check that the factors are actually greater than \(1\). In integer problems, signs and parity of factor pairs matter.
Mini-Checklist
Before computing, ask: is there a common factor? can the terms be grouped? is there a difference of squares or cubes? does the polynomial have a root? is the expression homogeneous? what happens when variables are equal?
Example 1. Common Factor and Trinomial
Even a simple common factor often reveals the next formula.
Problem. Factor \(6x^3y-9x^2y^2+3xy^3\).
First factor out \(3xy\):
\[6x^3y-9x^2y^2+3xy^3=3xy(2x^2-3xy+y^2).\]
Now \(2x^2-3xy+y^2=(2x-y)(x-y)\). Hence the answer is \(3xy(2x-y)(x-y)\).
Comment. The method has two steps: a common factor, then quadratic factorisation.
Example 2. Grouping
If a common factor is not visible at once, grouping can create it.
Problem. Factor \(ab+ac+bd+cd\).
Group the terms:
\[ab+ac+bd+cd=a(b+c)+d(b+c).\]
Now factor out \(b+c\), obtaining \((a+d)(b+c)\).
Comment. Good grouping makes an expression shorter, not longer.
Example 3. A Quadratic Trinomial with Parameters
The roots are sometimes visible without using the discriminant.
Problem. Factor \(x^2-(p+q)x+pq\).
We need two numbers whose sum is \(p+q\) and product is \(pq\). They are \(p\) and \(q\).
Therefore \(x^2-(p+q)x+pq=(x-p)(x-q)\).
Comment. This form often appears after a substitution or Vieta's formulas.
Example 4. The Factor Theorem
If a root of a polynomial is known, a linear factor is known as well.
Problem. Find \(k\) if \(x^3+kx^2-5x-2k\) is divisible by \(x-2\), and factor the polynomial for this value of \(k\).
Let \(P(x)=x^3+kx^2-5x-2k\). Divisibility by \(x-2\) means \(P(2)=0\).
\[P(2)=8+4k-10-2k=2k-2.\]
Thus \(k=1\). Then \(P(x)=x^3+x^2-5x-2=(x-2)(x^2+3x+1)\).
Comment. We did not guess the factorisation; we first forced the needed factor to appear.
Example 5. Sophie Germain's Trick
Sometimes a difference of squares must be created artificially.
Problem. Factor \(x^4+4y^4\).
Add and subtract \(4x^2y^2\):
\[x^4+4y^4=(x^2+2y^2)^2-(2xy)^2.\]
Therefore
\[x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2).\]
Comment. This is one of the main olympiad examples of hidden factorisation.
Example 6. The Cubic Identity
A symmetric expression in three variables often factors in a non-obvious way.
Problem. Factor \(a^3+b^3+c^3-3abc\).
Use the identity
\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).\]
It can be checked by expansion. In problems, the important point is to remember that the first factor is \(a+b+c\).
Comment. When \(a+b+c=0\), this identity immediately gives \(a^3+b^3+c^3=3abc\).
Example 7. A Homogeneous Expression
Homogeneity suggests factors of matching degree.
Problem. Factor \(a^4+a^2b^2+b^4\).
Add and subtract \(a^2b^2\):
\[a^4+a^2b^2+b^4=(a^2+b^2)^2-a^2b^2.\]
This is a difference of squares:
\[(a^2-ab+b^2)(a^2+ab+b^2).\]
Comment. Not every fourth-degree expression should be treated as a full expansion.
Example 8. Compositeness After a Shift
Factorisation helps not only to factor, but also to decide primality.
Problem. Find all integers \(n\) for which \((n+1)^4+4\) is prime.
Set \(m=n+1\). We need to understand when \(m^4+4\) is prime.
By Sophie Germain's trick, \(m^4+4=(m^2-2m+2)(m^2+2m+2)\).
If \(|m|>1\), this is a non-trivial factorisation. If \(m=0\), the number is \(4\), not prime. If \(m=1\) or \(m=-1\), the number is \(5\). Hence \(n=0\) or \(n=-2\).
Comment. A shift of variable keeps the same idea but hides the familiar structure.
Chapter
Equations and Systems
Key Idea
Olympiad equations and systems are rarely solved by applying a formula immediately. More often, one must first choose the right quantities: a sum, a product, a difference, symmetric expressions, or a new variable.
A good solution reduces the number of unknowns or turns the system into one equation.
Basic Facts
For the quadratic equation \(x^2-Sx+P=0\), the numbers \(S\) and \(P\) are the sum and product of the roots. In symmetric systems, introduce \(s=x+y\), \(p=xy\), and for three variables \(x+y+z\), \(xy+yz+zx\), \(xyz\).
The identities \(x^2+y^2=(x+y)^2-2xy\) and \(x^3+y^3=(x+y)^3-3xy(x+y)\) are especially useful.
When to Use This Method
Use substitutions when expressions repeat, when the system is symmetric, when \(x+\frac1x\), \(x^2+y^2\), or \(x^3+y^3\) appears, or when direct solving gives bulky computations.
In integer problems, transformations often lead to factorisation.
How to Recognise the Method
If the conditions do not change when \(x\) and \(y\) are swapped, look for \(x+y\) and \(xy\). If powers of one expression occur, try \(u=x^2\) or \(u=x+\frac1x\). If there is a parameter, first determine what it must guarantee: a root, tangency, uniqueness, or integrality.
Typical Mistakes
Do not lose solutions when squaring, dividing by an expression that may be zero, or substituting \(u=x^2\). After a substitution, always return to the original variables and check restrictions.
Mini-Checklist
Identify the type: quadratic equation, symmetric system, substitution, parameter, or integer solutions. Check the domain. Reduce the number of unknowns. Solve the new problem. Return to the original variables and verify the answers.
Example 1. A Quadratic via Sum and Product
The basic move is to see the sum and product without extra expansion.
Problem. Solve \(x^2-7x+12=0\).
We need two numbers with sum \(7\) and product \(12\). They are \(3\) and \(4\).
Thus \(x^2-7x+12=(x-3)(x-4)\), so \(x=3\) or \(x=4\).
Comment. This viewpoint will be needed in symmetric systems.
Example 2. A Biquadratic Equation
If only even powers occur, substitute \(x^2\).
Problem. Solve \(x^4-5x^2+4=0\).
Let \(u=x^2\), where \(u\ge0\). Then \(u^2-5u+4=0\), so \((u-1)(u-4)=0\).
Thus \(u=1\) or \(u=4\). Hence \(x=\pm1\) or \(x=\pm2\).
Comment. The condition \(u\ge0\) must be checked.
Example 3. A Symmetric System
A two-variable symmetric system is often solved through sum and product.
Problem. Find \(x,y\) if \(x+y=6\), \(x^2+y^2=20\).
From \(x^2+y^2=(x+y)^2-2xy\), we get \(20=36-2xy\), so \(xy=8\).
Then \(x,y\) are roots of \(t^2-6t+8=0\). Hence \(t=2\) or \(t=4\). Answer: \((2,4)\), \((4,2)\).
Comment. We recovered the pair as roots of a quadratic equation.
Example 4. Sum of Cubes in a System
Cubes should not be expanded separately.
Problem. Find \(x,y\) if \(x+y=5\), \(x^3+y^3=35\).
Let \(p=xy\). Then \(x^3+y^3=(x+y)^3-3xy(x+y)=125-15p\).
We have \(125-15p=35\), so \(p=6\). Now \(x,y\) are roots of \(t^2-5t+6=0\), giving \((x,y)=(2,3)\) or \((3,2)\).
Comment. The single parameter \(p=xy\) replaces a long expansion.
Example 5. Reciprocal Substitution
Expressions of the form x+1/x behave like a new variable.
Problem. If \(x+\frac1x=4\), find \(x^2+\frac1{x^2}\).
Square the equality:
\[\left(x+\frac1x\right)^2=x^2+2+\frac1{x^2}=16.\]
Therefore \(x^2+\frac1{x^2}=14\).
Comment. The condition \(x e0\) is already included in the expression.
Example 6. A Parameter as a Root Condition
A parameter is often chosen to create a specified root.
Problem. Find \(a\) if \(x=2\) is a root of \(x^2-ax+6=0\).
Substitute \(x=2\): \(4-2a+6=0\).
Thus \(10-2a=0\), so \(a=5\). The equation becomes \(x^2-5x+6=0\).
Comment. There is no need to solve the general equation first.
Example 7. Integer Solutions by Factoring
After transformation, an integer equation often becomes a product.
Problem. Find all integers \(x,y\) such that \(x^2-y^2=16\).
\((x-y)(x+y)=16\). The factors \(x-y\) and \(x+y\) have the same parity.
The suitable pairs are \((2,8)\), \((4,4)\), \((-2,-8)\), \((-4,-4)\), \((8,2)\), \((-8,-2)\). Solving \(x-y=u\), \(x+y=v\), we obtain all integer pairs: \((5,3)\), \((4,0)\), \((-5,-3)\), \((-4,0)\), \((5,-3)\), \((-5,3)\).
Comment. The parity of factor pairs must not be skipped.
Example 8. A System with Repeated Structure
Sometimes the solution is hidden in subtracting equations.
Problem. Solve the system \(x^2+y=6\), \(y^2+x=6\).
Subtract the second equation from the first:
\[x^2-y^2+y-x=0.\]
We get \((x-y)(x+y-1)=0\). If \(x=y\), then \(x^2+x=6\), so \(x=2\) or \(x=-3\). If \(x+y=1\), substitute \(y=1-x\): \(x^2+1-x=6\), so \(x^2-x-5=0\). Hence \(x=\frac{1\pm\sqrt{21}}2\), \(y=\frac{1\mp\sqrt{21}}2\).
Comment. Subtracting the system created a factorisation.
Chapter
Polynomials I
Key Idea
A polynomial can be studied not only through coefficients, but also through its values at points. Roots, remainders, equal values, and coefficient comparison often give a short solution.
Basic Facts
If \(P(a)=0\), then \(x-a\) divides \(P(x)\). When \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If two polynomials of degree at most \(n\) agree at \(n+1\) distinct points, then they are identical.
When to Use This Method
Use polynomial methods when values at several points are given, when roots are shared, when coefficients contain parameters, when coefficients are integers, or when expressions such as \(P(a)-P(b)\) occur. Polynomials are especially useful when direct expansion looks bulky.
How to Recognise the Method
The words “root”, “common root”, “values are equal”, “for all \(x\)”, and “for integers \(a,b\)” often point to the factor theorem, coefficient comparison, or the divisibility lemma \(a-b\mid P(a)-P(b)\).
Typical Mistakes
Do not conclude that two polynomials are identical from equality at one point. Do not forget the degree of a difference of polynomials. With integer coefficients, use integrality of values and divisibility of differences.
Mini-Checklist
Find the degree. Look at roots. Check special values. Consider the difference of two polynomials. If many values are known, compare the number of roots of the difference with its degree. If coefficients are integers, use \(a-b\mid P(a)-P(b)\).
Example 1. The Factor Theorem
A root immediately gives a linear factor.
Problem. Factor \(P(x)=x^3-4x^2+x+6\), knowing that \(P(2)=0\).
Since \(P(2)=0\), the polynomial is divisible by \(x-2\). Division gives \(P(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)\).
Comment. There is no need to guess all roots at once.
Example 2. Remainder
A value at a point replaces polynomial division.
Problem. Find the remainder when \(2x^4-x+5\) is divided by \(x-2\).
The remainder is \(P(2)=2\cdot16-2+5=35\).
Comment. This is the simplest form of the remainder theorem.
Example 3. Comparing Coefficients
A polynomial identity means equality of coefficients.
Problem. Find \(a,b\) if \(x^2+ax+b=(x-3)(x+5)\).
The right side is \(x^2+2x-15\). Hence \(a=2\), \(b=-15\).
Comment. Coefficient comparison is sometimes shorter than substitution.
Example 4. Difference of Polynomials
If two quadratic polynomials agree at three points, they agree everywhere.
Problem. Quadratic polynomials \(P\) and \(Q\) have equal values at \(x=0,1,2\). Prove that \(P(x)=Q(x)\) for all \(x\).
The polynomial \(R(x)=P(x)-Q(x)\) has degree at most \(2\) and three roots: \(0,1,2\). Therefore \(R\) is identically zero, so \(P=Q\).
Comment. This is one of the main tools of the module.
Example 5. Divisibility of Values
For an integer-coefficient polynomial, the difference of values is divisible by the difference of arguments.
Problem. Prove that if \(P(x)\) has integer coefficients, then \(P(10)-P(4)\) is divisible by \(6\).
For each \(k\), the number \(10^k-4^k\) is divisible by \(10-4=6\). Hence each difference of corresponding monomials is divisible by \(6\), so \(P(10)-P(4)\) is divisible by \(6\).
Comment. This lemma appears often in olympiad problems.
Example 6. A Common Root
A common root of two polynomials can be found from their difference.
Problem. The polynomials \(x^2+ax+b\) and \(x^2+bx+a\) have a common root, with \(a\ne b\). Find this root.
Let the common root be \(r\). Subtract the values: \(0=(r^2+ar+b)-(r^2+br+a)=(a-b)(r-1)\). Since \(a\ne b\), \(r=1\).
Comment. The difference removes the leading terms.
Example 7. Roots and Discriminant
For a monic quadratic, the discriminant is the square of the distance between the roots.
Problem. Let the roots of \(x^2+px+q\) be \(u\) and \(v\). Prove that the discriminant equals \((u-v)^2\).
By Vieta, \(p=-(u+v)\), \(q=uv\). Thus \(D=p^2-4q=(u+v)^2-4uv=(u-v)^2\).
Comment. This fact is needed when comparing quadratic trinomials.
Example 8. Substitution in an Identity
Sometimes one should replace \(x\) by \(-x\).
Problem. If \(F(x)=(x-a)(x-b)(x-c)-t\) has roots \(u,v,w\), express \((x+u)(x+v)(x+w)\) through \(a,b,c,t\).
Since \(F(x)=(x-u)(x-v)(x-w)\), substitute \(-x\): \((-x-a)(-x-b)(-x-c)-t=(-x-u)(-x-v)(-x-w)\). Multiplying by \(-1\), we get \((x+a)(x+b)(x+c)+t=(x+u)(x+v)(x+w)\).
Comment. This prepares for stronger problems about cubic roots.
Chapter
Sequences and Recurrences
Key Idea
A sequence in olympiad algebra is not just a list of numbers. One must understand the rule of change: differences, ratios, recurrence, invariant, or a hidden telescoping sum.
Basic Facts
For an arithmetic progression, \(a_n=a_1+(n-1)d\), \(S_n=\frac{n(a_1+a_n)}2\). For a geometric progression, \(a_n=a_1q^{n-1}\), \(S_n=a_1\frac{q^n-1}{q-1}\) for \(q\ne1\). Recurrences are often solved by substitution or induction.
When to Use This Method
Look for sequences when there is a repeated process, a transition from \(n\) to \(n+1\), a sum of many similar fractions, a maximum number after steps, or a condition “for every \(n\)”.
How to Recognise the Method
If neighbouring terms differ by a constant, it is arithmetic. If the ratio is constant, it is geometric. If the expression has the form \(u_k-u_{k+1}\), the sum telescopes. If a process preserves a quantity, look for an invariant.
Typical Mistakes
Do not confuse the index with the term. Do not forget the base case in induction. In recurrences with fractions, check the domain. In sequence-counting problems, first decide what can be chosen freely.
Mini-Checklist
Write the first terms. Inspect differences and ratios. Try a substitution such as \(b_n=a_n+c\) or \(b_n=\frac1{a_n}\). Check telescoping. Formulate the induction step.
Example 1. Arithmetic Progression
We start with the explicit formula.
Problem. In an arithmetic progression \(a_1=5\), \(d=3\). Find \(a_{20}\).
\(a_{20}=a_1+19d=5+57=62\).
Comment. Do not replace \(19\) by \(20\).
Example 2. Sum of a Progression
The sums of opposite pairs are constant.
Problem. Find \(7+10+13+\cdots+70\).
This is an arithmetic progression with \(a_1=7\), \(d=3\), \(a_n=70\). From \(7+3(n-1)=70\), \(n=22\). The sum is \(\frac{22(7+70)}2=847\).
Comment. First find the number of terms.
Example 3. Geometric Progression
A constant ratio gives a power.
Problem. In a geometric progression \(b_1=3\), \(q=2\). Find \(b_8\).
\(b_8=3\cdot2^7=384\).
Comment. Index \(8\) gives exponent \(7\).
Example 4. A Simple Recurrence
Sometimes the formula is guessed from the first terms and proved by induction.
Problem. The sequence is defined by \(a_1=1\), \(a_{n+1}=a_n+2n+1\). Prove that \(a_n=n^2\).
Base case: \(a_1=1=1^2\). If \(a_n=n^2\), then \(a_{n+1}=n^2+2n+1=(n+1)^2\). Thus the formula holds for all \(n\).
Comment. The difference of consecutive squares is \(2n+1\).
Example 5. A Linear Recurrence
A shift often turns a recurrence into a geometric one.
Problem. Let \(a_1=1\), \(a_{n+1}=2a_n+1\). Find \(a_n\).
Add \(1\): \(a_{n+1}+1=2(a_n+1)\). Hence \(a_n+1=2^n\), so \(a_n=2^n-1\).
Comment. The right substitution removes the constant term.
Example 6. Telescoping Sum
The sum opens after decomposing the fraction.
Problem. Find \(\sum_{k=1}^{n}\frac1{k(k+1)}\).
\(\frac1{k(k+1)}=\frac1k-\frac1{k+1}\). Therefore the sum is \(1-\frac1{n+1}=\frac n{n+1}\).
Comment. Almost all terms cancel.
Example 7. Invariant
A process may preserve not a term, but a combination of terms.
Problem. The sequence is defined by \(a_{n+1}=3a_n+2\), \(a_1=0\). Prove that \(a_n+1\) is a power of \(3\).
\(a_{n+1}+1=3a_n+3=3(a_n+1)\). Since \(a_1+1=1\), we get \(a_n+1=3^{n-1}\).
Comment. The invariant form is \(a_n+1\).
Example 8. Counting by Differences
Sometimes a sequence is determined by its start and its steps.
Problem. How many positive integer sequences \(a_1,\ldots,a_5\) have \(a_1=3\) and \(|a_{i+1}-a_i|\le1\)?
Each step is \(-1,0\), or \(1\), but we must ensure that no term becomes zero. Starting from \(3\), in four steps with decrease at most \(1\), reaching zero is impossible. Hence all \(3^4=81\) step choices work.
Comment. The difference sequence is the main object.
Chapter
Algebraic Transformations in Problems
Key idea
An algebraic transformation is useful when the original form hides a simpler structure: a square, a sum and product, homogeneity, symmetry, or a condition such as \(a+b+c=0\). The goal is not to expand everything, but to choose a form in which the problem becomes shorter.
In olympiad problems one often replaces several variables by a few expressions: \(x+y\), \(xy\), \(a+b+c\), \(ab+bc+ca\), \(abc\), \(\frac{x}{y}\), \(x+\frac{1}{x}\). After such a substitution the unnecessary details disappear.
Basic facts
Completing the square often uses \(x^2-2px=(x-p)^2-p^2\). For two variables, the formulas \(x^2+y^2=(x+y)^2-2xy\) and \(x^3+y^3=(x+y)^3-3xy(x+y)\) are very useful.
For three variables, symmetric expressions are essential. If \(a+b+c=0\), then \(a^2+b^2+c^2=-2(ab+bc+ca)\), \((a+b)(b+c)(c+a)=-abc\), and \[a^3+b^3+c^3=3abc.\]
If an expression is homogeneous, one may normalize one quantity, for example set \(b=1\), \(a+b+c=1\), or \(x+y=1\), provided this does not change the problem.
When to use this method
Use it when the problem contains symmetry, a sum of variables, a ratio of homogeneous expressions, expressions like \(x+\frac{1}{x}\), \(x^2+y^2\), \(a^3+b^3+c^3\), or when you need to prove that no solution exists.
Another signal is this: the statement has many letters, but the answer depends on only one or two combinations. Then try to reduce the number of variables.
How to recognise the method
First check whether the problem stays unchanged after multiplying all variables by the same number. If it does, look for a normalization. Then see whether the variables enter symmetrically: use \(x+y\), \(xy\), or \(a+b+c\), \(ab+bc+ca\), \(abc\).
If there is a condition \(a+b+c=0\), it is almost always worth substituting \(c=-a-b\), or directly applying the special identities for a zero sum.
Typical mistakes
It is risky to expand large brackets without a plan: the structure may disappear. Do not divide by an expression before checking that it is nonzero. When normalizing, make sure the chosen quantity is indeed positive or nonzero.
In rationality problems, the rationality of a sum does not imply the rationality of its terms. Usually one must find an additional expression that isolates one term.
Mini-checklist
1. Is there a square to complete? 2. Can \(x+y\) and \(xy\) be substituted? 3. Is the problem homogeneous? 4. Is there a zero-sum condition? 5. Is rationality hidden in reciprocals or in a sum of squares?
Example 1. Completing the square
This example shows how to find a minimum without calculus.
Problem. Find the least value of \(x^2-6x+y^2+4y+15\).
Complete the squares: \(x^2-6x=(x-3)^2-9\), \(y^2+4y=(y+2)^2-4\). Hence the expression equals \((x-3)^2+(y+2)^2+2\).
A sum of squares is nonnegative, so the minimum is \(2\). It is attained at \(x=3\), \(y=-2\).
Idea. If a quadratic expression asks for a minimum, first try completing squares.
Example 2. The zero-sum condition
The key is a special identity for \(a+b+c=0\).
Problem. Prove that if \(a+b+c=0\), then \(a^3+b^3+c^3=3abc\).
The identity \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\) immediately gives the result, because the first factor is zero.
Idea. A zero sum turns a sum of cubes into a product.
Example 3. Substituting the sum and product
A standard technique for symmetric systems in two variables.
Problem. Solve the system \(x+y=5\), \(x^2+y^2=13\).
Let \(s=x+y\), \(p=xy\). Then \(x^2+y^2=s^2-2p\), so \(13=25-2p\), hence \(p=6\).
The numbers \(x\) and \(y\) are the roots of \(t^2-5t+6=0\). Therefore \(\{x,y\}=\{2,3\}\).
Idea. A symmetric system often reduces to a quadratic equation.
Example 4. Homogeneity and normalization
This example teaches how to replace a ratio of variables by one new variable.
Problem. Let \(x,y>0\) and \(\frac{x}{y}+\frac{y}{x}=\frac{5}{2}\). Find the possible values of \(\frac{x-y}{x+y}\).
Put \(s=x+y\), \(p=xy\). Then \(\frac{x^2+y^2}{xy}=\frac{5}{2}\), that is \(\frac{s^2-2p}{p}=\frac{5}{2}\). Hence \(\frac{s^2}{p}=\frac{9}{2}\).
Now \(\left(\frac{x-y}{x+y}\right)^2=\frac{s^2-4p}{s^2}=1-\frac{4p}{s^2}=1-\frac{8}{9}=\frac{1}{9}\). Therefore \(\frac{x-y}{x+y}=\pm\frac{1}{3}\).
Idea. A homogeneous expression depends on the ratio, not on the scale.
Example 5. The hidden substitution \(x+\frac{1}{x}\)
Reciprocal quantities often need to be grouped before computation.
Problem. If \(x\neq 0\) and \(x+\frac{1}{x}=4\), find \(x^3+\frac{1}{x^3}\).
Use \(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)\). Thus the desired value is \(4^3-3\cdot4=52\).
Idea. There is no need to find \(x\); it is enough to work with the whole substitution.
Example 6. Reducing the number of variables
This example shows how three variables reduce to two symmetric sums.
Problem. Let \(a+b+c=0\) and \(a^2+b^2+c^2=12\). Find \(ab+bc+ca\).
Square \(a+b+c=0\): \(a^2+b^2+c^2+2(ab+bc+ca)=0\). Hence \(12+2(ab+bc+ca)=0\), so \(ab+bc+ca=-6\).
Idea. The sum of squares and the sum of pairwise products are connected by the square of the sum.
Example 7. Rationality through a sum of squares
The individual terms are not visible directly, but an additional identity isolates the needed expression.
Problem. Let \(\sin u+\sin v\) and \(\cos u+\cos v\) be rational. Prove that \(\cos(u-v)\) is rational.
Denote the given sums by \(S\) and \(C\). Then \(S^2+C^2\) is rational. On the other hand, \[S^2+C^2=2+2(\sin u\sin v+\cos u\cos v)=2+2\cos(u-v).\]
Therefore \(\cos(u-v)=\frac{S^2+C^2-2}{2}\) is rational.
Idea. Squaring a sum may reveal the needed angle difference.
Example 8. A strong form of zero sum
An olympiad step: view the numbers as roots of one polynomial.
Problem. Let \(a+b+c=0\), \(a^2+b^2+c^2=6\), \(a^3+b^3+c^3=6\). Find \(a,b,c\).
From the first condition and the sum of squares we get \(ab+bc+ca=-3\). By the zero-sum identity, \(a^3+b^3+c^3=3abc\), so \(abc=2\).
Thus \(a,b,c\) are the roots of \(t^3-3t-2\). Factor it: \(t^3-3t-2=(t-2)(t+1)^2\). Hence \(\{a,b,c\}=\{2,-1,-1\}\).
Idea. If three symmetric sums are known, the variables can be recovered as roots of a cubic polynomial.
Chapter
Introductory Inequalities
Key idea
An inequality in olympiad algebra is rarely proved by visual guessing. Usually it is reduced to an obvious nonnegative quantity: a square, a sum of squares, AM-GM, Cauchy-Schwarz, or a careful rearrangement of terms.
The main habit in this module is to check the equality case, not just the sign. Equality often suggests the right substitution and shows how sharp the estimate is.
Basic facts
For all real \(a,b\), \((a-b)^2\ge0\), hence \(a^2+b^2\ge2ab\). For positive numbers \(x_1,\ldots,x_n\), AM-GM says \[\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}.\]
A basic form of Cauchy-Schwarz is \[(a^2+b^2)(c^2+d^2)\ge(ac+bd)^2.\] In Engel form, one often uses \[\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}\] for \(a,b>0\).
When to use this method
AM-GM is natural when positive quantities have a fixed sum or product. Cauchy works well with sums of fractions like \(\frac{x^2}{a}\). Sums of squares appear when expanding leaves terms such as \((a-b)^2\).
If a problem asks for a maximum or minimum, first find the likely equality case, then choose an inequality whose equality occurs exactly there.
How to recognise the method
If the expression is symmetric in \(a,b,c\), try to reduce it to \((a-b)^2+(b-c)^2+(c-a)^2\). If there are positive fractions, check whether Cauchy in Engel form applies. If the product \(abc\) appears and the sum is fixed, AM-GM is usually nearby.
For rearrangement problems, sort two sequences in the same order. The difference between the “right” and “wrong” arrangement often becomes a product of two nonnegative differences.
Typical mistakes
AM-GM cannot be applied to negative numbers. In fractional inequalities, denominators must be positive. Multiplying an inequality by an expression of unknown sign may reverse the inequality and lose the solution.
Another common mistake is proving a weaker inequality and not noticing that it does not imply the target. Always check numerically whether the chosen estimate is strong enough.
Mini-checklist
1. Are all quantities positive? 2. What is the equality case? 3. Can the difference be written as a sum of squares? 4. Is there a form \(\sum \frac{x^2}{a}\)? 5. Should the variables be sorted?
Example 1. A square of a difference
The basic idea is to prove an inequality through obvious nonnegativity.
Problem. Prove that \(a^2+b^2\ge2ab\) for all real \(a,b\).
Since \((a-b)^2\ge0\), we have \(a^2-2ab+b^2\ge0\). Hence \(a^2+b^2\ge2ab\). Equality holds only when \(a=b\).
Example 2. AM-GM for two numbers
This example shows how to find a minimum of an expression with a reciprocal.
Problem. For \(x>0\), prove that \(x+\frac{1}{x}\ge2\).
By AM-GM, \(x+\frac{1}{x}\ge2\sqrt{x\cdot\frac{1}{x}}=2\). Equality holds at \(x=1\).
Example 3. Fixed sum
If the sum of positive numbers is fixed, the product is often bounded above.
Problem. Let \(a,b,c>0\) and \(a+b+c=6\). Prove that \(abc\le8\).
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Hence \(2\ge\sqrt[3]{abc}\), so \(abc\le8\). Equality holds at \(a=b=c=2\).
Example 4. Cauchy in Engel form
Engel form turns a sum of fractions into the square of a sum.
Problem. For \(x,y>0\), prove that \(\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\).
By Cauchy, \(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{(x+y)^2}{x+y}=x+y\). Equality holds when \(x=y\).
Example 5. Sum of reciprocals
Cauchy often proves a lower bound for reciprocals.
Problem. Let \(x,y,z>0\) and \(x+y+z=1\). Prove that \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\).
By Cauchy, \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\ge(1+1+1)^2=9\). Since \(x+y+z=1\), the claim follows. Equality holds when \(x=y=z=\frac{1}{3}\).
Example 6. Nesbitt's inequality
This is a standard first example where Cauchy is stronger than direct addition.
Problem. For \(a,b,c>0\), prove that \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\).
Write \(\frac{a}{b+c}=\frac{a^2}{a(b+c)}\). By Cauchy, the sum is at least \(\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}\). The denominator is \(2(ab+bc+ca)\), and \((a+b+c)^2\ge3(ab+bc+ca)\). Thus the result is \(\frac{3}{2}\).
Example 7. Rearrangement intuition
When two sets are ordered in the same way, the sum of paired products is largest.
Problem. If \(a\le b\) and \(x\le y\), prove that \(ay+bx\le ax+by\).
The difference between the right and left sides is \(ax+by-ay-bx=(b-a)(y-x)\ge0\). Therefore \(ay+bx\le ax+by\).
Example 8. Equality as a guide
A strong solution usually explains immediately when equality can occur.
Problem. Let \(x,y,z>0\) and \(xyz=1\). Prove that \((1+x)(1+y)(1+z)\ge8\).
By AM-GM, \(1+x\ge2\sqrt{x}\), \(1+y\ge2\sqrt{y}\), \(1+z\ge2\sqrt{z}\). Multiplying, \((1+x)(1+y)(1+z)\ge8\sqrt{xyz}=8\). Equality holds when \(x=y=z=1\).
Chapter
Introductory Functional Equations
Key idea
A functional equation is an equation where the unknown is a function rather than a number. At the start one almost always substitutes special values: \(0\), \(1\), \(x\), \(-x\), \(x=y\), \(y=0\). These substitutions give the first anchor values and properties of the function.
The main goal of this introductory module is to avoid guessing the function immediately. Instead, extract information step by step: find \(f(0)\), \(f(1)\), check injectivity or surjectivity, and then move toward a linear or additive structure.
Basic facts
If \(f(x+y)=f(x)+f(y)\), one usually first gets \(f(0)=0\). On \(\mathbb Z\) and \(\mathbb Q\), additivity implies \(f(n)=nf(1)\) and \(f\left(\frac{p}{q}\right)=\frac{p}{q}f(1)\).
On \(\mathbb R\), additivity alone is not enough to conclude \(f(x)=cx\). Additional regularity is needed: monotonicity, boundedness on an interval, nonnegativity on positive numbers, or a similar condition.
When to use this method
Substitution is almost always needed. Injectivity is useful when \(f(A)=f(B)\) can give \(A=B\). Surjectivity is useful when \(f(y)\) needs to be replaced by an arbitrary number. Testing a linear form is useful when \(f(x)=ax+b\) is expected but must still be justified.
How to recognise the method
If the equation contains \(x+y\), try \(y=0\) and \(x=0\). If it contains \(f(f(x))\), look for injectivity or invertibility. If it contains \(x+f(y)\), first fix \(y\), then choose \(y\) so that \(f(y)\) becomes a convenient argument.
Typical mistakes
Do not assume the function is linear without proof. Do not divide by \(f(x)\) before proving this value is nonzero. Do not use “for every \(t\) there is \(y\) with \(f(y)=t\)” before surjectivity has been proved.
Mini-checklist
1. Have \(0\) and \(1\) been substituted? 2. Are \(f(0)\), \(f(1)\) known? 3. Is there injectivity or surjectivity? 4. Can one pass to \(g(x)=f(x)+c\) or \(g(x)=f(x)-x^2\)? 5. Have the found functions been checked in the original equation?
Example 1. Finding \(f(0)\)
The first move is to substitute the simplest argument.
Problem. Let \(f(x+y)=f(x)+f(y)\) for all integers \(x,y\). Prove that \(f(0)=0\).
Substitute \(x=0\), \(y=0\): \(f(0)=f(0)+f(0)\). Hence \(f(0)=0\).
Example 2. Additivity on integers
On \(\mathbb Z\), additivity is completely determined by the value at one.
Problem. If \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)\), \(f(1)=3\), find \(f(n)\).
For \(n>0\), \(f(n)=nf(1)=3n\). Also \(0=f(0)=f(n+(-n))=f(n)+f(-n)\), so \(f(-n)=-3n\). Answer: \(f(n)=3n\) for all \(n\in\mathbb Z\).
Example 3. Substituting \(x=0\)
Sometimes one value immediately determines the recurrence.
Problem. Let \(f(n+1)=f(n)+2n+1\) for \(n\ge0\), \(f(0)=0\). Find \(f(n)\).
Add the equalities for \(n=0,1,\ldots,k-1\): \(f(k)=1+3+\cdots+(2k-1)=k^2\). Hence \(f(n)=n^2\).
Example 4. Testing a linear function
A linear form may be tested, but it cannot simply be assumed without conditions.
Problem. Find all linear functions \(f(x)=ax+b\) such that \(f(x+y)=f(x)+f(y)+4\).
Substitute \(ax+b\). We get \(a(x+y)+b=ax+b+ay+b+4\). The coefficients of \(x,y\) match automatically, and the constant terms give \(b=2b+4\), so \(b=-4\). Answer: \(f(x)=ax-4\), where \(a\) is any real number.
Example 5. Injectivity
Injectivity allows one to remove \(f\) from both sides of an equality.
Problem. Let \(f\) be injective and \(f(x+1)=f(y+1)\). Prove that \(x=y\).
By injectivity, \(x+1=y+1\). Therefore \(x=y\).
Example 6. Surjectivity
Surjectivity lets us replace a function value by an arbitrary number.
Problem. Let \(f:\mathbb R\to\mathbb R\) be surjective. Prove that there exists \(t\) such that \(f(t)=0\).
By surjectivity, every real number is a value of the function. In particular, \(0\) is a value: there exists \(t\) such that \(f(t)=0\).
Example 7. Additivity on \(\mathbb Q\)
On rational numbers, the Cauchy equation does not require regularity.
Problem. Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(1)=5\). Find \(f\left(\frac{7}{3}\right)\).
From \(3\cdot\frac{1}{3}=1\), we get \(3f\left(\frac{1}{3}\right)=f(1)=5\), hence \(f\left(\frac{1}{3}\right)=\frac{5}{3}\). Then \(f\left(\frac{7}{3}\right)=7f\left(\frac{1}{3}\right)=\frac{35}{3}\).
Example 8. Substitution for a quadratic term
Sometimes one subtracts a known part to obtain ordinary additivity.
Problem. Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)+2xy\), \(f(1)=1\). Find \(f(x)\).
Set \(g(x)=f(x)-x^2\). Then \(g(x+y)=f(x+y)-(x+y)^2=f(x)-x^2+f(y)-y^2=g(x)+g(y)\). Also \(g(1)=0\). On \(\mathbb Q\), an additive function with \(g(1)=0\) is zero. Hence \(f(x)=x^2\).
Chapter
Algebraic Number Problems
Key idea
Algebraic problems with integers are usually solved not by blind search, but by changing the form: factorization, passing to the sum and product of roots, working modulo a number, or bounding possible divisors.
This module connects algebra with number theory: an identity gives divisibility, Vieta's formulas give integer restrictions, and modular arithmetic quickly rules out impossible equations.
Basic facts
The product of several consecutive integers is often divisible by a factorial: for example, \(n(n+1)\) is divisible by \(2\), and \(n(n+1)(n+2)\) is divisible by \(6\).
If an integer \(r\) is a root of a monic polynomial with integer coefficients, then \(r\) divides the constant term. For a quadratic equation with integer roots, Vieta's formulas are useful: the sum of the roots is \(-b\), and their product is \(c\).
Squares modulo \(4\) give only residues \(0\) and \(1\), and squares modulo \(3\) also give only \(0\) and \(1\). This often proves that integer solutions are impossible.
When to use this method
If an equation contains \(x^2-y^2\), look for a difference of squares. If there is a rational expression with integer variables, try to transform it into a product such as \((x-a)(y-b)=N\). If a polynomial must have an integer root, first check the divisors of the constant term.
How to recognise the method
The words “integers”, “positive integers”, “divisible by”, and “find all solutions” almost always mean that the possibilities must be restricted. Divisibility restricts factors, modular arithmetic restricts residues, and Vieta turns unknown roots into a sum and product.
Typical mistakes
Do not check only positive divisors if variables may be negative. Do not cancel \(a\) from \(ab\mid ac\) without considering \(a=0\). In modular problems, all possible residues must be checked, not just one example.
Mini-checklist
1. Can the expression be factored? 2. Which residues are possible modulo \(3\), \(4\), \(5\)? 3. If there is an integer root, which divisors must be checked? 4. Can Vieta's formulas be applied? 5. Have negative solutions been included?
Example 1. Product of neighboring integers
Problem. Prove that \(n^3-n\) is divisible by \(6\) for every integer \(n\).
Factor: \(n^3-n=n(n-1)(n+1)\). This is the product of three consecutive integers. One of them is divisible by \(3\), and at least one is divisible by \(2\), so the whole product is divisible by \(6\).
Example 2. Integer root of a polynomial
Problem. Find the integer roots of \(x^3-2x^2-5x+6\).
An integer root must divide \(6\), so we check \(\pm1,\pm2,\pm3,\pm6\). The roots are \(1,2,-3\). Hence \(x^3-2x^2-5x+6=(x-1)(x-2)(x+3)\).
Example 3. Vieta over integers
Problem. Integers \(x,y\) satisfy \(x+y=9\), \(xy=20\). Find \(x,y\).
The numbers \(x,y\) are roots of \(t^2-9t+20=0\). The factorization \(t^2-9t+20=(t-4)(t-5)\) gives \(\{x,y\}=\{4,5\}\).
Example 4. A fraction becomes a product
Problem. Find positive integers \(x,y\) such that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{5}\).
Multiply by \(5xy\): \(5x+5y=xy\). Thus \(xy-5x-5y=0\), so \((x-5)(y-5)=25\). Checking positive divisors of \(25\), we get \((x,y)=(6,30),(10,10),(30,6)\).
Example 5. Modular check
Problem. Prove that \(x^2+1=3y\) has no integer solutions.
Modulo \(3\), a square has residue \(0\) or \(1\). Hence \(x^2+1\) has residue \(1\) or \(2\), while \(3y\) has residue \(0\). Contradiction.
Example 6. Difference of squares
Problem. Find integer solutions of \(x^2-y^2=21\).
\((x-y)(x+y)=21\). The factors \(x-y\) and \(x+y\) have the same parity, and \(21\) is odd, so both are odd. Checking factor pairs \(1\cdot21\), \(3\cdot7\), and the negative pairs, we get \((x,y)=(11,10),(11,-10),(5,2),(5,-2),(-11,-10),(-11,10),(-5,-2),(-5,2)\).
Example 7. A quadratic form
Problem. Find integers \(x,y\) such that \(x^2-xy+y^2=1\).
Use \(4(x^2-xy+y^2)=(2x-y)^2+3y^2\). Then \((2x-y)^2+3y^2=4\), so \(y\in\{-1,0,1\}\). Checking gives \((1,0),(-1,0),(0,1),(0,-1),(1,1),(-1,-1)\).
Example 8. First descent
Problem. Prove that \(x^2+y^2=5xy\) has no positive integer solutions.
Assume a solution exists and choose one with minimal \(x+y\). Let \(x\ge y\). Consider the equation as a quadratic in \(x\): \(X^2-5yX+y^2=0\). The other root is \(x'=5y-x\), it is an integer, and since the product of the roots is \(y^2\), \(x'=\frac{y^2}{x}>0\). If \(x>y\), then \(x'
Chapter
Mixed Algebra Problems I
Key idea
In a mixed module the method is not announced in advance. The student must recognize the form of the problem: factorization, sum-product substitution, modular arithmetic, inequality, functional equation, or recurrence.
The main strategy is to look for structure before computing. A good first minute: check symmetry, integrality, completing the square, the presence of \(x+y\), \(xy\), \(f(0)\), \(f(1)\), fixed sum, or fixed product.
Basic facts
A symmetric system in two variables often reduces to \(s=x+y\), \(p=xy\). A difference of squares gives \((x-y)(x+y)\). For integer roots of a monic polynomial, first check divisors of the constant term. For inequalities, \((a-b)^2\ge0\), AM-GM, and Cauchy are useful.
In functional equations, begin with special values. In sequences, look for finite differences, telescoping, or an explicit formula from the first terms.
When to use this method
If the problem asks to “find all”, you need a mechanism that restricts the possibilities. If it asks to “prove”, look for an identity or a nonnegative form. If the problem looks bulky, try replacing the variables by more stable expressions.
How to recognise the method
Integers suggest divisibility and residues. Quadratic expressions suggest completing the square or Vieta. Positive variables with a fixed sum suggest AM-GM or Cauchy. An equation containing \(f(x+y)\) almost always starts with \(x=0\) or \(y=0\).
Typical mistakes
In mixed problems it is easy to apply a favorite method without checking. Do not expand everything if factors can be kept. Always verify the found answers in the original condition, especially in parameter and integer problems.
Mini-checklist
1. Is there symmetry? 2. Is there integrality or a modular obstruction? 3. Can a square be completed? 4. Is there a natural substitution? 5. Have equality cases, signs, and negative solutions been checked?
Example 1. The method is not announced
Problem. Solve \(x+y+xy=19\), \(x^2+y^2=25\).
Let \(s=x+y\), \(p=xy\). Then \(s+p=19\), \(s^2-2p=25\). Since \(p=19-s\), \(s^2+2s-63=0\), so \(s=7\) or \(s=-9\). For \(s=7\), \(p=12\), the roots are \(3\) and \(4\). For \(s=-9\), \(p=28\), the discriminant is negative. Answer: \((3,4),(4,3)\).
Example 2. Hidden divisibility
Problem. Prove that \(n^3+3n^2+2n\) is divisible by \(6\).
Factor: \(n^3+3n^2+2n=n(n+1)(n+2)\). This is the product of three consecutive integers, so it is divisible by \(6\).
Example 3. Parameter through Vieta
Problem. For which integers \(a\) does \(x^2+ax+18\) have integer roots?
Let the roots be \(r,s\). Then \(rs=18\), \(r+s=-a\). The possible sums of factor pairs of \(18\) are \(\pm19,\pm11,\pm9\). Hence \(a\in\{-19,-11,-9,9,11,19\}\).
Example 4. Inequality without a hint
Problem. Let \(x,y,z>0\). Prove \(\frac{x^2}{2x+y}+\frac{y^2}{2y+z}+\frac{z^2}{2z+x}\ge\frac{x+y+z}{3}\).
By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{(2x+y)+(2y+z)+(2z+x)}=\frac{(x+y+z)^2}{3(x+y+z)}=\frac{x+y+z}{3}\).
Example 5. Functional equation
Problem. Let \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)+2mn\), \(f(1)=2\). Find \(f(n)\).
Set \(g(n)=f(n)-n^2\). Then \(g(m+n)=g(m)+g(n)\), and \(g(1)=1\). Hence \(g(n)=n\), so \(f(n)=n^2+n\).
Example 6. Modular obstruction
Problem. Prove that \(x^2+y^2+z^2=8k+7\) has no integer solutions.
A square modulo \(8\) is \(0,1,4\). A sum of three such residues cannot be \(7\) modulo \(8\). But the right-hand side is congruent to \(7\). Contradiction.
Example 7. Sequence
Problem. \(u_0=1\), \(u_1=3\), \(u_{n+2}=3u_{n+1}-2u_n\). Prove that \(u_n=2^{n+1}-1\).
The check for \(n=0,1\) is true. If the formula holds for \(n,n+1\), then \(u_{n+2}=3(2^{n+2}-1)-2(2^{n+1}-1)=2^{n+3}-1\). The induction is complete.
Example 8. Primality of an expression
Problem. Find all integers \(n\) for which \(n^4+4\) is prime.
Use Sophie Germain's factorization: \(n^4+4=(n^2-2n+2)(n^2+2n+2)\). For \(|n|>1\), both factors are greater than \(1\), so the number is composite. For \(n=1\) and \(n=-1\), we get \(5\), which is prime. For \(n=0\), we get \(4\). Answer: \(n=\pm1\).
Chapter
Strategy Notes
Key idea
Strategy in algebra begins not with computation, but with choosing a viewpoint. The same problem may look like a system, a polynomial, an inequality, or a divisibility problem; the right form shortens the solution sharply.
Basic facts
A sum and product suggest Vieta. A zero sum suggests \(a^3+b^3+c^3=3abc\). A fixed sum in an inequality suggests the equality case. Integers suggest divisors and residues. Functional equations almost always begin with \(0\), \(1\), or \(x=y\).
When to use this method
Use substitution when the same combinations repeat. Look for factorization when there is a difference of powers or a constant term. Use symmetry when variables enter in the same way. Look at polynomial degree when values at several points are given.
How to recognise the method
Ask: what stays unchanged if variables are swapped? which values are obvious? where can equality occur? which residues are possible? If the answer depends on only two combinations, reduce the number of variables.
Typical mistakes
The most common mistake is expanding before recognizing structure. The second is finding an answer and not checking it. The third is using a heavy theorem where one substitution or factorization is enough.
Mini-checklist
1. Is there a natural substitution? 2. Are there factors? 3. Is there symmetry? 4. Does polynomial degree matter? 5. Where is the equality case? 6. Is there a quick modular obstruction?
Example 1. When to substitute
Problem. Solve \(x+y=11\), \(x^2+y^2=65\).
The expressions \(x+y\) and \(x^2+y^2\) are visible, so find \(xy\): \(65=121-2xy\), hence \(xy=28\). Thus \(x,y\) are roots of \(t^2-11t+28=0\), namely \(4\) and \(7\).
Example 2. When to factor
Problem. Find integer solutions of \(x^2-y^2=27\).
The difference of squares gives \((x-y)(x+y)=27\). Checking odd factor pairs gives \((14,\pm13),(6,\pm3)\) and the corresponding negatives in \(x\).
Example 3. When to use symmetry
Problem. If \(a+b+c=0\), \(a^2+b^2+c^2=14\), \(a^3+b^3+c^3=18\), find \(a,b,c\).
We have \(ab+bc+ca=-7\), and \(3abc=18\), so \(abc=6\). The numbers are roots of \(t^3-7t-6=0=(t-3)(t+1)(t+2)\). Answer: permutations of \((3,-1,-2)\).
Example 4. When to look at degree
Problem. \(P\) has degree at most \(2\), \(P(0)=2\), \(P(1)=6\), \(P(2)=12\). Find \(P(3)\).
The first differences are \(4,6\), so the second difference is \(2\), and the next first difference is \(8\). Hence \(P(3)=20\).
Example 5. When to seek the equality case
Problem. For \(a+b+c=3\), \(a,b,c>0\), prove \(ab+bc+ca\le3\).
Equality is expected at \(a=b=c=1\). Use \((a+b+c)^2\ge3(ab+bc+ca)\), which gives \(9\ge3(ab+bc+ca)\).
Example 6. When to work modulo
Problem. Prove that \(x^2+y^2=4k+3\) is impossible in integers.
A square modulo \(4\) is \(0\) or \(1\), so a sum of two squares is \(0,1,2\), but not \(3\).
Example 7. When to substitute \(0\)
Problem. \(f(x+y)=f(x)+f(y)+2\). Find \(f(0)\).
At \(x=y=0\), \(f(0)=2f(0)+2\), so \(f(0)=-2\).
Example 8. When to look at differences
Problem. \(u_0=2\), \(u_{n+1}-u_n=4n+1\). Find \(u_n\).
\(u_n=2+\sum_{k=0}^{n-1}(4k+1)=2+2n(n-1)+n=2n^2-n+2\).
Chapter
Mock Olympiads I
Key idea
A mock olympiad tests method choice, not a single topic. Each variant contains an equation or factorization problem, a polynomial or sequence problem, an inequality, and a functional or mixed problem.
Basic facts
Before solving, classify the problem quickly: factors, substitution, Vieta, finite differences, AM-GM/Cauchy, substitution in a functional equation, or a modular check.
When to use this method
If the method is not visible after 2–3 minutes, write down the structure: repeated expressions, symmetric variables, integrality, and possible equality cases.
How to recognise the method
A system with \(x+y\), \(xy\) asks for Vieta. Several polynomial values ask for differences. Positive variables ask for inequalities. \(f(x+y)\) asks for \(0\) and \(1\).
Typical mistakes
Do not stay too long on one problem. In a mock variant, collect solvable ideas first, then return to the hard problems.
Mini-checklist
1. Is the method chosen? 2. Is the answer checked? 3. Are all integer cases included? 4. Is the equality case found? 5. Is the solution written without gaps?
Example 1. Factorisation
Problem. Solve \(x^2-9x+20=0\).
\((x-4)(x-5)=0\), hence \(x=4\) or \(x=5\).
Example 2. Differences
Problem. \(P(0)=1\), \(P(1)=4\), \(P(2)=9\), and \(\deg P\le2\). Find \(P(3)\).
The first differences are \(3,5\), so the next one is \(7\). Therefore \(P(3)=16\).
Example 3. AM-GM
Problem. For \(x>0\), prove \(x+\frac{16}{x}\ge8\).
By AM-GM, \(x+\frac{16}{x}\ge2\sqrt{16}=8\).
Example 4. Cauchy equation on \(\mathbb Q\)
Problem. \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(2)=10\). Find \(f\left(\frac{3}{5}\right)\).
\(f(1)=5\), hence \(f(q)=5q\). The answer is \(3\).
Example 5. Modular check
Problem. Prove that \(x^2+y^2=4k+3\) is impossible in integers.
Modulo \(4\), a square is only \(0\) or \(1\), so a sum of two squares cannot be \(3\).
Example 6. Sequence
Problem. \(u_0=0\), \(u_{n+1}-u_n=2n+1\). Find \(u_n\).
\(u_n=1+3+\cdots+(2n-1)=n^2\).
Example 7. Vieta
Problem. \(x+y=12\), \(xy=35\). Find \(x,y\).
The numbers are roots of \(t^2-12t+35=0\), so they are \(5\) and \(7\).
Example 8. Descent
Problem. Why does \(x^2+y^2=5xy\) have no positive integer solutions?
In a minimal solution, the other root \(5y-x=\frac{y^2}{x}\) produces a smaller positive integer solution. This is a contradiction.