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Suppose that \((a^n-b^n)|(a^n+b^n)\). We may assume that \(a > b\). Then there is an integer \(k > 1\) such that \(a^n+b^n = k(a^n-b^n)\) which gives \((k+1)b^n = (k-1)a^n\). Suppose \(p>2\) is a prime such that \(p|(k+1)\). Since \((k+1,k-1) = (k+1,2) \leqslant 2\) then \(p \nmid (k-1)\). Looking a...

We have that \[ \binom{3n}{n} = \frac{1}{2\pi i}\oint_C \frac{(1+z)^{3n}}{z^{n+1}} \, dz\] where $C$ is any simple closed curve enclosing the origin. [This follows by the residue theorem as the coefficient of $z^n$ in $(1+z)^{3n}$ is $\binom{3n}{n}$.] In what follows, we will take $C$ to be the unit...

Just observe that $\displaystyle{\frac{1}{n!}, \frac{2}{n!}, \ldots, \frac{n}{n!}}$ are all distinct terms of the sequence.

Hi Apostole. The right word is "order" not "provision".

I know the answer but it is too nice to spoil this problem by giving it. Let's hope that somebody who does not know the answer will work it out.

Let \(q_1,q_2,\ldots\) be an enumeration of the rationals. Suppose we have already defined \(g,h\) on \(q_1,\ldots,q_k\) such that their sum is equal to \(f\) on those points and they are 1 to 1 so far. When defining \(g\) and \(h\) on \(q_{k+1}\) it is enough to pick any \(q \in \mathbb{Q}\) such t...

Why try something complicated while something simple would do? Maybe you wanted to demand a strict inequality? In that case my simple example does not work any more but the non-continuous solutions of Cauchy's equation do work. However there are still simpler functions which also work. For example \...

The function which is equal to $1$ everywhere except at $0$ on which it is equal to $0$ is such a function.

We know that the sum of all terms is \[ \frac{n(n+1)(2n+1)}{6}\] and thus their average is \[ \frac{(n+1)(2n+1)}{6}\] Since $2$ must divide $(n+1)(2n+1)$ we must have $n = 2m-1$ for some natural number $m$. So we need $\displaystyle{ \frac{m(4m-1)}{3}}$ to be a perfect square. Since $m$ and $4m-1$ a...

Well, \[\binom{2n}{n} = \frac{(2n)!}{n!n!}.\] We have that $p|(2n!)$ as $p < 2n$, but $p \nmid (n!)^2$ as $p$ is prime and $p > n$. So $\binom{2n}{n} \equiv 0 \bmod p$.

Let us write \(a_n\) for the \(n\)-th term of the series and \(s_n\) for the partial sum of the first \(n\) terms. Observe that \[ s_{6n} = \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{8n-1} \right) - \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{4n} \right) = H_{8n} - \frac{1}{2}H_...