Chapter
Theory
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.
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 \(x=y\).
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.
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.
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.
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?
Examples
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 \(7\).
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 \(x\).
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 \((3,-1,-2)\).
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 \(P(3)=20\).
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 \(9\ge3(ab+bc+ca)\).
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 \(3\).
Problem. \(f(x+y)=f(x)+f(y)+2\). Find \(f(0)\).
At \(x=y=0\), \(f(0)=2f(0)+2\), so \(f(0)=-2\).
Problem. \(u_0=2\), \(u_{n+1}-u_n=4n+1\). Find \(u_n\).
\(u_n=2+\sum_{k=0}^{n-1}(4k+1)=2+2n(n-1)+n=2n^2-n+2\).
Problems
Solve \(x^2-6x+8=0\).
Look for factors of the constant term.
\(x^2-6x+8=(x-2)(x-4)\), so \(x=2\) or \(x=4\).
Find the minimum of \(x^2-2x+3\).
Complete the square.
\(x^2-2x+3=(x-1)^2+2\). The minimum is \(2\).
Prove that \(n^2-n\) is even for every integer \(n\).
Factor into neighboring terms.
\(n^2-n=n(n-1)\). Among two consecutive integers, one is even.
\(f(0)=3\), \(f(n+1)=f(n)+1\). Find \(f(4)\).
Take four steps.
\(f(4)=3+4=7\).
Find \(x,y\) if \(x+y=5\), \(xy=6\).
View them as roots of a quadratic.
\(x,y\) are roots of \(t^2-5t+6=0\), hence \(\{x,y\}=\{2,3\}\).
Let \(a+b+c=0\). Prove \(a^3+b^3+c^3=3abc\).
Use the factorization of \(a^3+b^3+c^3-3abc\).
\(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\).
Find the integer roots of \(x^3-3x^2-4x+12\).
Try grouping.
\(x^3-3x^2-4x+12=x^2(x-3)-4(x-3)=(x-3)(x-2)(x+2)\). The roots are \(3,2,-2\).
For \(x>0\), prove \(x+\frac{9}{x}\ge6\).
Equality is expected at \(x=3\).
By AM-GM, \(x+\frac{9}{x}\ge2\sqrt{9}=6\).
\(P\) has degree at most \(2\), \(P(0)=2\), \(P(1)=5\), \(P(2)=10\). Find \(P(3)\).
Look at second differences.
The first differences are \(3,5\), so the second difference is \(2\). The next first difference is \(7\), hence \(P(3)=17\).
Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(5)=20\). Find \(f\left(\frac{3}{2}\right)\).
Find \(f(1)\).
\(5f(1)=20\), so \(f(1)=4\). Therefore \(f(q)=4q\), and \(f\left(\frac{3}{2}\right)=6\).
Solve \(x+y+xy=29\), \(x^2+y^2=41\).
Set \(s=x+y\), \(p=xy\).
\(s+p=29\), \(s^2-2p=41\). We get \(s^2+2s-99=0\), so \(s=9\) or \(s=-11\). For \(s=9\), \(p=20\), \(\{x,y\}=\{4,5\}\); the second case gives no real roots.
Find positive integers \(x,y\) if \((x-4)(y-4)=16\).
Check positive divisors of \(16\).
\((x-4,y-4)\) equals \((1,16),(2,8),(4,4),(8,2),(16,1)\). Hence \((x,y)=(5,20),(6,12),(8,8),(12,6),(20,5)\).
Let \(x,y,z>0\). Prove \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{x+y+z}{2}\).
Add the denominators.
By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{(x+y)+(y+z)+(z+x)}=\frac{x+y+z}{2}\).
\(u_0=1\), \(u_{n+1}-u_n=2n+3\). Find \(u_n\).
Sum the differences.
\(u_n=1+\sum_{k=0}^{n-1}(2k+3)=1+n(n-1)+3n=n^2+2n+1=(n+1)^2\).
Prove that \(x^2+y^2=4k+3\) is impossible in integers.
Work modulo \(4\).
Squares modulo \(4\) are \(0\) or \(1\). A sum of two squares cannot have residue \(3\).
A polynomial \(P\) satisfies \(P(x+1)-P(x)=2x+1\), \(P(0)=0\). Find \(P(n)\) for integers \(n\ge0\).
Sum the equalities from \(0\) to \(n-1\).
\(P(n)-P(0)=\sum_{k=0}^{n-1}(2k+1)=n^2\). Hence \(P(n)=n^2\).
Let \(a,b,c>0\), \(a+b+c=3\). Prove \(a^2+b^2+c^2\ge3\).
Equality is expected at \(a=b=c=1\).
By Cauchy, \((a^2+b^2+c^2)(1+1+1)\ge(a+b+c)^2=9\). Hence \(a^2+b^2+c^2\ge3\).
Find all integer solutions of \(x^2+y^2=2xy+16\).
Move \(2xy\) to the left.
\((x-y)^2=16\), so \(x-y=\pm4\). All solutions are \((t+4,t)\) and \((t,t+4)\), where \(t\in\mathbb Z\).
Prove that \(x^2+y^2=7xy\) has no positive integer solutions.
Choose a minimal solution and replace the larger root by the other root of the quadratic equation.
Assume a solution exists, \(x\ge y\), with minimal \(x+y\). The case \(x=y\) is impossible. Consider \(X^2-7yX+y^2=0\). The other root \(x'=7y-x=\frac{y^2}{x}\) is positive, integral, and smaller than \(y\). Then \((x',y)\) is a smaller positive solution. Contradiction.
Find all real triples \((a,b,c)\) such that \(a+b+c=0\), \(a^2+b^2+c^2=14\), \(a^3+b^3+c^3=18\).
Find \(ab+bc+ca\) and \(abc\), then form the polynomial.
From the first two conditions, \(ab+bc+ca=-7\). By the zero-sum identity, \(3abc=18\), so \(abc=6\). Thus \(a,b,c\) are roots of \(t^3-7t-6=0=(t-3)(t+1)(t+2)\). Answer: all permutations of \((3,-1,-2)\).
Ladders