48th Austrian Mathematical Olympiad

Answer: The two solutions are $(0,0)$ and $(0,2)$.

Since $2017^a$ is always odd, $b$ must be even, so $b=2c$, $c$ integer. Therefore, $2017^a=64(c^6-c)+1$ and thus $2017^a\equiv 1(\mathrm{mod}64)$. But we find $2017\equiv 33(\mathrm{mod}64)$ and $2017^2\equiv (1+32{)}^2=1+2\cdot 32+32^2\equiv 1(\mathrm{mod}64)$, so that the powers of $2017$ modulo $64$ alternate between $1$ and $33$. Therefore, $a$ is even and $2017^a$ is a perfect square. We denote the polynomial on the right-hand side of the given equation by $r(b)=b^6-32b+1$ and show that it lies between two consecutive squares for $b>4$:

Let $b>4$. We have $r(b)<b^6=(b^3{)}^2$ for $b>0$. On the other hand, $r(b)>(b^3-1{)}^2$ because $b^6-32b+1>b^6-2b^3+1\iff b>4$. Since the square $2017^a$ is now between two consecutive squares, there are no solutions in this case.

Since $b$ is even, it remains to check $b=4$, $b=2$ and $b=0$.

For $b=4$, we regard the equation modulo $3$ and get $1\equiv 1-2+1=0$, therefore, there is no solution in this case.

For $b=2$, we get $2017^a=2^6-2^6+1$, so we get the solution $(a,b)=(0,2)$.

For $b=0$, we get $2017^a=1$, so we get the solution $(a,b)=(0,0)$.

Therefore, $(0,0)$ and $(0,2)$ are the only solutions.