Let \(n\) be a natural number and \(x=2n^2+n\). Prove that the sum of the square of the \(n+1\) consecutive integers starting at \(x\) is the sum of the square of the \(n\) consecutive integers starting at \(x+n+1\).
For example, when \(n=2\), we have \(10^2+11^2+12^2=13^2+14^2\)!
Find the contrapositive of the statement: “If in every school there is a class with at least \(20\) students, then there is a school with at least \(10\) students".
Show that if \(a\) and \(b\) are numbers, then \(a^2-b^2=(a-b)\times (a+b)\).
Show that if \(x,y,z\) are distinct nonzero numbers such that \(x+y+z = 0\), then we have \[\left(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}\right)\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right) = 9.\]
The Chinese remainder theorem is a fundamental result in number theory that allows one to decompose congruence problems to into simpler ones. The theorem says the following.
Suppose that \(m_1,m_2\) are coprime (i.e: they have no prime factors in common) natural numbers and \(a_1,a_2\) are integers. Then there is a unique integer \(x\) in the range \(0\leq x \leq m_1m_2-1\) such that \[x \equiv a_1 \pmod{m_1} \quad \text{ and } \quad x \equiv a_2 \pmod{m_2},\] where the notation \(x\equiv y \pmod{z}\) means that \(x-y=kz\) for some integer \(k\). Prove the Chinese remainder theorem using the pigeonhole principle.
We have ten positive integers \(x_1,\dots,x_{10}\) such that \(10\leq x_i\leq 99\) for \(1\leq i\leq 10\). Prove that there are two disjoint subsets of \(x_1,\dots,x_{10}\) with equal sums of their elements.
Let \(P(x)\) be a polynomial with integral coefficients. Suppose there exist four distinct integers \(a,b,c,d\) with \(P(a) = P(b) = P(c) = P(d) = 5\). Prove that there is no integer \(k\) with \(P(k) = 8\).
For which natural number \(n\) is the polynomial \(1+x^2+x^4+\dots+x^{2n-2}\) divisible by the polynomial \(1 +x+x^2+\dots+x^{n-1}\)?
Let \(P(x)\) be a polynomial with integer coefficients. Set \(P^1(x) = P(x)\) and \(P^{i+1}(x) = P(P^i(x))\). Show that if \(t\) is an integer such that \(P^k(t)=t\) for some natural number \(k\), then in fact we have \(P^2(t) = t\).
(IMO 2006) Let \(P(x)\) be a polynomial of degree \(n > 1\) with integer coefficients and let \(k\) be a positive integer. Consider the polynomial \(Q(x) = P^k(x)\). Prove that there are at most \(n\) integers \(t\) such that \(Q(t) = t\).