Practice
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: \(n=\pm1\).