Chapter
Theory
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.
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.
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.
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.
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.
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.
Examples
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.
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.
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.
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.
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.
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.
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.
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.
Problems
Solve \(x^2-9x+20=0\).
Find two numbers with sum \(9\) and product \(20\).
\(x^2-9x+20=(x-4)(x-5)\). Hence \(x=4\) or \(x=5\).
Solve \(x^4-10x^2+9=0\).
Set \(u=x^2\), remembering that \(u\ge0\).
Let \(u=x^2\). Then \(u^2-10u+9=0\), so \(u=1\) or \(u=9\).
Thus \(x=\pm1\) or \(x=\pm3\).
Find all pairs \(x,y\) such that \(x+y=8\), \(xy=15\).
Treat \(x,y\) as roots of a quadratic equation.
The numbers \(x,y\) are roots of \(t^2-8t+15=0\).
Thus \((t-3)(t-5)=0\). Therefore \((x,y)=(3,5)\) or \((5,3)\).
Find \(a\) if \(x=3\) is a root of \(x^2-ax+12=0\).
Substitute \(x=3\) into the equation.
We get \(9-3a+12=0\), so \(21-3a=0\).
Therefore \(a=7\).
Let \(x\ne0\) and \(x+\frac1x=5\). Find \(x^2+\frac1{x^2}\).
Square the equality.
\(\left(x+\frac1x\right)^2=x^2+2+\frac1{x^2}=25\).
Hence \(x^2+\frac1{x^2}=23\).
Find \(x,y\) if \(x+y=7\), \(x^2+y^2=25\).
First find \(xy\).
From \(x^2+y^2=(x+y)^2-2xy\), we get \(25=49-2xy\), hence \(xy=12\).
Then \(x,y\) are roots of \(t^2-7t+12=0\), so \(t=3\) or \(t=4\). Answer: \((3,4)\), \((4,3)\).
Solve \(x^4-13x^2+36=0\).
Set \(u=x^2\), then solve the quadratic equation in \(u\).
We get \(u^2-13u+36=0\), so \((u-4)(u-9)=0\).
Since \(u=x^2\), we have \(x^2=4\) or \(x^2=9\). Hence \(x=\pm2,\pm3\).
Solve the system \(x^2+y=12\), \(y^2+x=12\).
Subtract one equation from the other.
Subtracting gives \(x^2-y^2+y-x=0\), so \((x-y)(x+y-1)=0\).
If \(x=y\), then \(x^2+x=12\), so \(x=3\) or \(x=-4\). This gives \((3,3)\), \((-4,-4)\).
If \(x+y=1\), then \(y=1-x\), and \(x^2+1-x=12\). Thus \(x^2-x-11=0\), so \(x=\frac{1\pm3\sqrt5}{2}\), \(y=\frac{1\mp3\sqrt5}{2}\).
Find all integers \(x,y\) such that \(x^2-y^2=21\).
Factor the left side and list factor pairs of \(21\).
\((x-y)(x+y)=21\). The two factors have the same parity, and the product is odd, so both are odd.
The pairs \((1,21)\), \((3,7)\), \((-1,-21)\), \((-3,-7)\) give \((11,10)\), \((5,2)\), \((-11,-10)\), \((-5,-2)\).
For which \(a\) does \(x^2-4x+a=0\) have exactly one real root?
A quadratic equation has exactly one real root when its discriminant is zero.
The discriminant is \(D=16-4a\). We need \(D=0\), so \(16-4a=0\).
Thus \(a=4\).
Find \(x,y\) if \(x+y=6\), \(x^3+y^3=72\).
Express the sum of cubes through \(xy\).
\(x^3+y^3=(x+y)^3-3xy(x+y)=216-18xy\).
From \(216-18xy=72\), we get \(xy=8\). Then \(x,y\) are roots of \(t^2-6t+8=0\), so \((x,y)=(2,4)\), \((4,2)\).
Real numbers \(x,y,z\) satisfy \(x+y+z=6\), \(x^2+y^2+z^2=12\). Prove that \(xy+yz+zx=12\).
Square \(x+y+z=6\).
\((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\).
Substitute the data: \(36=12+2(xy+yz+zx)\). Hence \(xy+yz+zx=12\).
Let \(x\ne0\) and \(x+\frac1x=3\). Find \(x^3+\frac1{x^3}\).
Use \(u^3\), where \(u=x+\frac1x\).
Let \(u=x+\frac1x=3\). Then \(u^3=x^3+\frac1{x^3}+3\left(x+\frac1x\right)\).
Thus \(27=x^3+\frac1{x^3}+9\), so \(x^3+\frac1{x^3}=18\).
Find all integers \(a\) for which \(x^2-ax+12=0\) has two integer roots.
The roots must be integer divisors of \(12\).
Let the roots be \(m,n\). Then \(mn=12\), \(m+n=a\).
List integer divisor pairs: \((1,12)\), \((2,6)\), \((3,4)\), and the corresponding negative pairs. Thus \(a=13,8,7,-13,-8,-7\).
Solve the system \(x+y=5\), \(x^2-y^2=15\).
Factor \(x^2-y^2\).
\(x^2-y^2=(x-y)(x+y)\). Since \(x+y=5\), we get \(5(x-y)=15\), so \(x-y=3\).
Solving \(x+y=5\), \(x-y=3\), gives \(x=4\), \(y=1\).
Solve \( (x^2-3x)^2-2(x^2-3x)-8=0 \).
Set \(u=x^2-3x\).
Let \(u=x^2-3x\). Then \(u^2-2u-8=0\), so \(u=4\) or \(u=-2\).
If \(x^2-3x=4\), then \(x^2-3x-4=0\), so \(x=4\) or \(x=-1\). If \(x^2-3x=-2\), then \(x^2-3x+2=0\), so \(x=1\) or \(x=2\).
Real numbers \(x,y,z\) satisfy \(x+y+z=3\), \(x^2+y^2+z^2=3\). Prove that \(x=y=z=1\).
Consider \((x-1)^2+(y-1)^2+(z-1)^2\).
\((x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3=3-6+3=0\).
A sum of squares is zero only when all squares are zero. Hence \(x=y=z=1\).
Solve the system \(x^2+2y=9\), \(y^2+2x=9\).
Subtract the equations and factor the result.
Subtract: \(x^2-y^2+2y-2x=0\), so \((x-y)(x+y-2)=0\).
If \(x=y\), then \(x^2+2x=9\), so \(x=-1\pm\sqrt{10}\). This gives two pairs \((-1+\sqrt{10},-1+\sqrt{10})\), \((-1-\sqrt{10},-1-\sqrt{10})\).
If \(x+y=2\), then \(y=2-x\). Substitution gives \(x^2+4-2x=9\), so \(x^2-2x-5=0\). Hence \(x=1\pm\sqrt6\), \(y=1\mp\sqrt6\).
Find all positive integers \(x,y\) such that \(xy=x+y+5\).
Move everything to the left and add \(1\) to get a product.
We have \(xy-x-y=5\). Add \(1\): \((x-1)(y-1)=6\).
Since \(x,y\) are positive integers, \(x-1\) and \(y-1\) are positive divisors of \(6\). The pairs \((1,6)\), \((2,3)\), \((3,2)\), \((6,1)\) give \((x,y)=(2,7),(3,4),(4,3),(7,2)\).
Find all real \(x,y\) if \(x+y=2\) and \(x^4+y^4=2\).
First express everything through \(p=xy\).
Let \(p=xy\). Then \(x^2+y^2=4-2p\).
\(x^4+y^4=(x^2+y^2)^2-2x^2y^2=(4-2p)^2-2p^2=16-16p+2p^2\).
Since this equals \(2\), \(16-16p+2p^2=2\), so \(p^2-8p+7=0\). Thus \(p=1\) or \(p=7\). But real \(x,y\) with \(x+y=2\) require \(p\le1\). Hence \(p=1\), and \(x,y\) are roots of \(t^2-2t+1=0\). Therefore \(x=y=1\).
Find all integers \(a\) for which \(x^2-(a+1)x+a+6=0\) has root \(x=3\) or \(x=4\).
Substitute \(x=3\) and \(x=4\) separately.
For \(x=3\): \(9-3(a+1)+a+6=0\), so \(12-2a=0\), hence \(a=6\).
For \(x=4\): \(16-4(a+1)+a+6=0\), so \(18-3a=0\), hence \(a=6\). Thus the only value is \(a=6\). For this value, both \(3\) and \(4\) are indeed roots.
Real numbers \(x,y\) satisfy \(x^2+y^2=2x+4y-5\). Prove that \(x=1\), \(y=2\).
Move everything to the left and complete squares.
Rewrite as \(x^2-2x+y^2-4y+5=0\).
This is \((x-1)^2+(y-2)^2=0\). A sum of squares is zero only when \(x=1\), \(y=2\).
Find all real triples \(x,y,z\) such that \(x+y+z=3\), \(xy+yz+zx=3\), \(xyz=1\).
Treat \(x,y,z\) as roots of a cubic polynomial.
The numbers \(x,y,z\) are roots of \(t^3-3t^2+3t-1\).
But \(t^3-3t^2+3t-1=(t-1)^3\). Therefore all three roots equal \(1\), so \(x=y=z=1\).
Find all positive integers \(x,y,z\) such that \(xyz=x+y+z+2\).
Assume \(x\le y\le z\) and bound \(x\).
Without loss of generality, let \(x\le y\le z\). If \(x\ge3\), then \(xyz\ge3yz\), while the right side is at most \(3z+2\); for \(y\ge3\), \(z\ge3\), this is impossible. Hence \(x=1\) or \(x=2\).
If \(x=1\), then \(yz=y+z+3\), so \((y-1)(z-1)=4\). This gives \((y,z)=(2,5),(3,3),(5,2)\).
If \(x=2\), then \(2yz=y+z+4\). For \(y=1\), \(z=5\); for \(y=2\), \(z=2\); for \(y\ge3\), the left side is too large: \(2yz-y-z-4\ge6z-3-z-4=5z-7>0\). Taking permutations into account, all solutions are permutations of \((1,2,5)\), \((1,3,3)\), and \((2,2,2)\).
Solve the system \(x+y+xy=11\), \(x^2+y^2=25\).
Set \(s=x+y\), \(p=xy\).
Let \(s=x+y\), \(p=xy\). Then \(s+p=11\), and \(x^2+y^2=s^2-2p=25\).
Since \(p=11-s\), we get \(s^2-2(11-s)=25\), so \(s^2+2s-47=0\). Thus \(s=-1\pm4\sqrt3\), \(p=11-s\).
Real \(x,y\) require \(s^2-4p\ge0\), which holds for both values. Then \(x,y\) are roots of \(t^2-st+p=0\). Thus for each \(s=-1\pm4\sqrt3\), \(p=11-s\), the pairs \((x,y)\) are the two permutations of the roots of \(t^2-st+p=0\).
Positive numbers \(x,y,z\) have the following property: the values \(x+2y^2+2z^2\), \(y+2z^2+2x^2\), \(z+2x^2+2y^2\) are equal. Must \(x=y=z\)?
Subtract two equal expressions. A factor \(x-y\) should appear.
Answer: no, not necessarily. We show how to find a counterexample.
From the equality of the first two expressions, \(x+2y^2=y+2x^2\). Move all terms to one side:
\[x-y-2(x^2-y^2)=0.\]
Thus \((x-y)(1-2(x+y))=0\). Therefore for each pair of variables, either they are equal or their sum is \(\frac12\).
Take \(x=y=\frac16\), \(z=\frac13\). Then \(x+y=\frac13\), but \(x+z=y+z=\frac12\). Check:
\[x+2y^2+2z^2=\frac16+\frac1{18}+\frac2{9}=\frac49,\]
and the other two expressions are also equal to \(\frac49\). However, \(x,y,z\) are not all equal. Hence the statement is false.
Ladders