Indonesia, 2013

[Fatin Farhan]

An integer $$n$$ is called $$\text{"elephantine"}$$ if there exists a positive
integer $$x$$ such that $$x^{nx} + 1$$ is divisible by $$2^n$$.
$$1. \text{ Prove } 2015 \text{ is Elephantine. }$$
$$2. \text{ Find the smallest } x\text{ such that }x^{nx} + 1\text{ is divisible by }2^n$$ for $$n = 2015.$$

[mutasimmim]

One line solution by LTE if I'm not missing something.

[Nirjhor]

As \(x\) is odd, \(2\mid x+1\). Also then \(2015x\) is odd. As \(2\nmid 1\) and \(2\nmid x\), we have \[v_2\left(x^{2015x}+1\right)=v_2(x+1)+v_2(2015x)=v_2(x+1).\]So we need the smallest \(x\) such that \(2^{2015}\mid x+1\), which clearly is \(\boxed{2^{2015}-1.}\)

In fact the same LTE bash generalizes to the fact that any odd \(n\) is elephantine and the smallest \(x\) for that \(n\) is \(2^n-1\).