SAUDI ARABIAN MATHEMATICAL COMPETITIONS

It is clear that $x=1$ is a solution of the problem (with $y$ any positive integer).

We now show that for $x>1$, the given equation has no solution. In fact, suppose that there are $x>1$ and $y$ are positive integers satisfying the equation. Then, one has $y>1$ and $$x^{2016}+x^{2015}+\cdots +x+1=y^{2015}-1=(y-1)(y^{2014}+y^{2013}+\cdots +y+1).$$ Put $p=2017$, and note that $p$ is a prime. Let $q$ be a prime factor of $x^{p-1}+\cdots +x+1$. Then $(x,q)=1$ and by Fermat's little theorem we get $q\mid x^{q-1}-1$. But $q\mid x^p-1$, it follows that the order of $x$ modulo $q$ is a divisor of $(p,q-1)$ which is either $1$ or $p$. This shows that $q\mid x-1$ or $p\mid q-1$.

If $q\mid x-1$ then $x^{p-1}+\cdots +x+1\equiv p(\mathrm{mod}q)$, i.e. $q\mid p$, this gives $p=q$. This means that any prime divisor of $x^{p-1}+\cdots +x+1$ is congruent either to $0$ or to $1(\mathrm{mod}p)$, hence any positive divisor of $x^{p-1}+\cdots +x+1$ is too. In particular, $y-1$ and $y^{2014}+y^{2013}+\cdots +y+1$ (which are positive divisors of $x^{p-1}+\cdots +x+1$) are congruent either to $0$ or to $1(\mathrm{mod}p)$.

But, if $p\mid y-1$ then $$y^{2014}+y^{2013}+\cdots +y+1\equiv 2015\not\equiv 0,1(\mathrm{mod}p),$$ we get a contradiction.

Hence, $y-1\equiv 1(\mathrm{mod}p)$, i.e. $y\equiv 2(\mathrm{mod}p)$, then $$y^{2014}+y^{2013}+\cdots +y+1\equiv 2^{2014}+2^{2013}+\cdots +2+1\equiv 2^{2015}-1(\mathrm{mod}p).$$ This shows that $2^{2015}-1\equiv 0(\mathrm{mod}p)$ or $2^{2015}-1\equiv 1(\mathrm{mod}p)$. That is, either $p\mid 2^{2015}-1$ or $p\mid 2^{2014}-1$. On the other hand, by Fermat's little theorem again, $p\mid 2^{p-1}-1=2^{2016}-1$. From this, we easily get $p\mid 1$ or $p\mid 3$ which is impossible. The problem is therefore solved.