IMO 2006 Shortlisted Problems

The equation has no integer solutions. To show this, we first prove a lemma.

Lemma. If $x$ is an integer and $p$ is a prime divisor of $\frac{x^7-1}{x-1}$ then either $p\equiv 1(\mathrm{mod}7)$ or $p=7$.

Proof. Both $x^7-1$ and $x^{p-1}-1$ are divisible by $p$, by hypothesis and by Fermat's little theorem, respectively. Suppose that $7$ does not divide $p-1$. Then $\gcd (p-1,7)=1$, so there exist integers $k$ and $m$ such that $7k+(p-1)m=1$. We therefore have $$x\equiv x^{7k+(p-1)m}\equiv {\left(x^7\right)}^k\cdot {\left(x^{p-1}\right)}^m\equiv 1(\mathrm{mod}p),$$ and so $$\frac{x^7-1}{x-1}=1+x+\cdots +x^6\equiv 7(\mathrm{mod}p).$$ It follows that $p$ divides $7$, hence $p=7$ must hold if $p\equiv 1(\mathrm{mod}7)$ does not, as stated. $\square$

The lemma shows that each positive divisor $d$ of $\frac{x^7-1}{x-1}$ satisfies either $d\equiv 0(\mathrm{mod}7)$ or $d\equiv 1(\mathrm{mod}7)$.

Now assume that $(x,y)$ is an integer solution of the original equation. Notice that $y-1>0$, because $\frac{x^7-1}{x-1}>0$ for all $x\ne 1$. Since $y-1$ divides $\frac{x^7-1}{x-1}=y^5-1$, we have $y\equiv 1(\mathrm{mod}7)$ or $y\equiv 2(\mathrm{mod}7)$ by the previous paragraph. In the first case, $1+y+y^2+y^3+y^4\equiv 5(\mathrm{mod}7)$, and in the second $1+y+y^2+y^3+y^4\equiv 3(\mathrm{mod}7)$. Both possibilities contradict the fact that the positive divisor $1+y+y^2+y^3+y^4$ of $\frac{x^7-1}{x-1}$ is congruent to $0$ or $1$ modulo $7$. So the given equation has no integer solutions.