Chapter
Theory
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.
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.
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.
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 \(y=0\).
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.
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?
Examples
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: \((3,4),(4,3)\).
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 \(6\).
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 \(a\in\{-19,-11,-9,9,11,19\}\).
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 \(\frac{(x+y+z)^2}{(2x+y)+(2y+z)+(2z+x)}=\frac{(x+y+z)^2}{3(x+y+z)}=\frac{x+y+z}{3}\).
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 \(f(n)=n^2+n\).
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.
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.
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: \(n=\pm1\).
Problems
Find the least value of \(x^2-10x+29\).
Complete the square.
\(x^2-10x+29=(x-5)^2+4\). The minimum is \(4\), attained at \(x=5\).
Solve the system \(x+y=8\), \(xy=15\).
View \(x,y\) as roots of a quadratic equation.
\(x,y\) are roots of \(t^2-8t+15=0=(t-3)(t-5)\). Answer: \((3,5),(5,3)\).
Prove that \(n(n+1)(n+2)\) is divisible by \(6\) for every integer \(n\).
Among three consecutive integers there is a multiple of \(3\) and an even number.
Among three consecutive integers, one is divisible by \(3\), and one by \(2\). Therefore the product is divisible by \(2\cdot3=6\).
Let \(f(0)=5\) and \(f(n+1)=f(n)+2\) for \(n\ge0\). Find \(f(4)\).
Make four transitions.
\(f(1)=7\), \(f(2)=9\), \(f(3)=11\), \(f(4)=13\).
For \(x>0\), prove that \(x+\frac{4}{x}\ge4\).
Apply AM-GM to \(x\) and \(\frac{4}{x}\).
\(x+\frac{4}{x}\ge2\sqrt{x\cdot\frac{4}{x}}=4\). Equality occurs at \(x=2\).
Solve \(x+y+xy=19\), \(x^2+y^2=25\).
Set \(s=x+y\), \(p=xy\).
We get \(s+p=19\), \(s^2-2p=25\). Then \(s^2+2s-63=0\), so \(s=7\) or \(s=-9\). For \(s=7\), \(p=12\), so \(\{x,y\}=\{3,4\}\). For \(s=-9\), \(p=28\), there are no real roots. Answer: \((3,4),(4,3)\).
Find all integers \(a\) for which \(x^2+ax+18\) has integer roots.
Use that the product of the roots is \(18\).
Let the roots be \(r,s\). Then \(rs=18\), \(r+s=-a\). Possible sums of factor pairs are \(\pm19,\pm11,\pm9\). Therefore \(a\in\{-19,-11,-9,9,11,19\}\).
Let \(a+b+c=0\). Prove that \(a^3+b^3+c^3=3abc\).
Use the identity for the sum of cubes.
\(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\). Hence \(a^3+b^3+c^3=3abc\).
Find all linear functions \(f(x)=ax+b\) such that \(f(x+y)=f(x)+f(y)+3\).
Substitute \(ax+b\).
\(a(x+y)+b=ax+b+ay+b+3\). The constants give \(b=2b+3\), so \(b=-3\), while \(a\) is arbitrary. Answer: \(f(x)=ax-3\).
Let \(u_0=0\), \(u_{n+1}=u_n+3n+1\). Find \(u_n\).
Sum \(3k+1\) from \(0\) to \(n-1\).
\(u_n=\sum_{k=0}^{n-1}(3k+1)=3\frac{n(n-1)}{2}+n=\frac{3n^2-n}{2}\).
Find all integer solutions of \(x^2-y^2=35\).
Factor as \((x-y)(x+y)\) and keep track of the order of factors.
Let \(u=x-y\), \(v=x+y\). Then \(uv=35\), \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{2}\). Both factors are odd, so all divisor pairs work. From \((u,v)=(1,35),(35,1),(5,7),(7,5)\) and the corresponding negative pairs, we get \((18,17),(18,-17),(6,1),(6,-1),(-18,-17),(-18,17),(-6,-1),(-6,1)\).
A polynomial \(P(x)\) of degree at most \(2\) satisfies \(P(0)=1\), \(P(1)=3\), \(P(2)=7\). Find \(P(3)\).
For a quadratic polynomial, second differences are constant.
The first differences are \(2\) and \(4\), so the second difference is \(2\). The next first difference is \(6\). Therefore \(P(3)=7+6=13\).
Find positive integer solutions of \(\frac{1}{x}+\frac{1}{y}=\frac{1}{8}\).
Transform it to \((x-8)(y-8)=64\).
From \(8x+8y=xy\), we get \((x-8)(y-8)=64\). Positive divisors of \(64\) are \(1,2,4,8,16,32,64\). Thus the solutions are \((9,72),(10,40),(12,24),(16,16)\) and the symmetric pairs.
Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(3)=12\). Find \(f\left(\frac{5}{2}\right)\).
First find \(f(1)\).
\(f(3)=3f(1)=12\), so \(f(1)=4\). Then \(f(q)=4q\) for rational \(q\), hence \(f\left(\frac{5}{2}\right)=10\).
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove that \(ab+bc+ca\le3\).
Use \((a+b+c)^2\ge3(ab+bc+ca)\).
\((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge3(ab+bc+ca)\). Since \(a+b+c=3\), \(9\ge3(ab+bc+ca)\), so \(ab+bc+ca\le3\).
Find all real triples \((a,b,c)\) such that \(a+b+c=0\), \(a^2+b^2+c^2=8\), \(a^3+b^3+c^3=0\).
Use \(a^3+b^3+c^3=3abc\).
From the zero sum, \(a^3+b^3+c^3=3abc\), so \(abc=0\). Also \(ab+bc+ca=-4\). If one number is zero, say \(c=0\), then \(a+b=0\), \(a^2+b^2=8\), hence \(a=\pm2\), \(b=\mp2\). All solutions are permutations of \((2,-2,0)\).
Find all integer triples \((x,y,z)\) such that \(x^2+y^2+z^2=xy+yz+zx+3\).
Multiply by \(2\) and use the sum of squared differences.
We get \((x-y)^2+(y-z)^2+(z-x)^2=6\). The integer squares must be \(1,1,4\). Therefore the differences are \(1,1,-2\), up to signs and order. Hence the triple consists of three consecutive integers. All solutions are permutations of \((t-1,t,t+1)\), \(t\in\mathbb Z\).
Find all integer pairs \((a,b)\) for which \(x^2+ax+b\) has two integer roots differing by \(1\).
Let the roots be \(n\) and \(n+1\).
Let the roots be \(n\) and \(n+1\), where \(n\in\mathbb Z\). By Vieta, \(a=-(2n+1)\), \(b=n(n+1)\). Conversely, for such \(a,b\), the polynomial is \((x-n)(x-n-1)\). Answer: \((a,b)=(-(2n+1),n(n+1))\), \(n\in\mathbb Z\).
Let \(u_0=1\), \(u_1=3\), \(u_{n+2}=3u_{n+1}-2u_n\). Prove that \(u_n=2^{n+1}-1\).
Check the base cases and make the induction step.
For \(n=0,1\), the formula is true. If \(u_n=2^{n+1}-1\) and \(u_{n+1}=2^{n+2}-1\), then \(u_{n+2}=3(2^{n+2}-1)-2(2^{n+1}-1)=2^{n+3}-1\). The formula is proved.
Let \(x,y,z>0\). Prove \[\frac{x^2}{2x+y}+\frac{y^2}{2y+z}+\frac{z^2}{2z+x}\ge\frac{x+y+z}{3}.\]
Add the denominators and apply Cauchy.
By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{3(x+y+z)}=\frac{x+y+z}{3}\).
Let \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)+2mn\), \(f(1)=2\). Find \(f(n)\).
Subtract \(n^2\).
Let \(g(n)=f(n)-n^2\). Then \(g(m+n)=g(m)+g(n)\), and \(g(1)=1\). Hence \(g(n)=n\), so \(f(n)=n^2+n\).
Prove that \(x^2+y^2+z^2=8k+7\) has no integer solutions.
Squares modulo \(8\) are \(0,1,4\).
Possible square residues modulo \(8\) are \(0,1,4\). A sum of three such residues is never \(7\) modulo \(8\). The right-hand side is congruent to \(7\), contradiction.
Prove that \(x^2+y^2=6xy\) has no positive integer solutions.
Choose a solution with minimal sum and view the equation as a quadratic in the larger variable.
Assume a positive solution exists and choose one with minimal \(x+y\). Let \(x>y\). As a quadratic in \(x\), the equation is \(X^2-6yX+y^2=0\). The other root \(x'=6y-x=\frac{y^2}{x}\) is positive, integral, and smaller than \(y\). Then \((x',y)\) is a smaller positive solution, contradiction.
Find all integers \(n\) for which \(n^4+4\) is prime.
Use the factorization \(n^4+4=(n^2-2n+2)(n^2+2n+2)\).
We have \(n^4+4=(n^2-2n+2)(n^2+2n+2)\). For \(|n|>1\), both factors are greater than \(1\), so the number is composite. For \(n=0\), we get \(4\). For \(n=1\) and \(n=-1\), we get \(5\), which is prime. Answer: \(n=\pm1\).
Ladders