MMO2025 Round 2

Suppose there exist integers $x$ and $y$ such that $x^4+x^3+x^2+x=y^6+61$.

Let us consider the equation modulo $7$.

First, note that $y^6\equiv 0$ or $1(\mathrm{mod}7)$ for any integer $y$, since by Fermat's Little Theorem, $y^6\equiv 1$ if $y$ is not divisible by $7$, and $0$ otherwise.

So $y^6\equiv 0$ or $1(\mathrm{mod}7)$.

Now, $61\equiv 5(\mathrm{mod}7)$, so $y^6+61\equiv 5$ or $6(\mathrm{mod}7)$.

Therefore, $x^4+x^3+x^2+x\equiv 5$ or $6(\mathrm{mod}7)$.

Let us compute $x^4+x^3+x^2+x(\mathrm{mod}7)$ for all $x=0,1,2,3,4,5,6$:

- $x=0$: $0^4+0^3+0^2+0=0$ - $x=1$: $1+1+1+1=4$ - $x=2$: $16+8+4+2=30\equiv 2(\mathrm{mod}7)$ - $x=3$: $81+27+9+3=120\equiv 1(\mathrm{mod}7)$ - $x=4$: $256+64+16+4=340\equiv 4(\mathrm{mod}7)$ - $x=5$: $625+125+25+5=780\equiv 3(\mathrm{mod}7)$ - $x=6$: $1296+216+36+6=1554\equiv 1(\mathrm{mod}7)$

So the possible residues are $0,1,2,3,4$ modulo $7$.

But $y^6+61\equiv 5$ or $6(\mathrm{mod}7)$, which are not among the possible residues for $x^4+x^3+x^2+x$.

Therefore, there are no integer solutions to the equation.