Practice
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 \((3,-1,-2)\).