Proving an equation has no roots in integers.

Question

Prove$$[a]^3+{[a+1]}^3+{[a+2]}^3+{[a+3]}^3+{[a+4]}^3+{[a+5]}^3+{[a+6]}^3=[b]^4+{[b+1]}^4$$ Has no integral solutions of $a,b$.

It is from my Olympiad and I truly have no idea how to proceed on this. I would welcome any hints for this problem.

Hint: try using modulo $7$

Answer 1

Note that $\pmod 7$ the list $\quad a,(a+1),(a+2),(a+3),(a+4),(a+5),(a+6)$

Is exactly $\quad 0,1,2,3,4,5,6\quad $ is some order so:

$\begin{align}&a^3+(a+1)^3+(a+2)^3+(a+3)^3+(a+4)^3+(a+5)^3+(a+6)^3\\ &\equiv 0^3+1^3+2^3+3^3+4^3+5^3+6^3\\ &\equiv 0^3+1^3+2^3+3^3+(-3)^3+(-2)^2+(-1)^3\\ &\equiv 0^3+(1^3-1^3)+(2^3-2^3)+(3^3-3^3)\\ &\equiv 0\pmod 7\end{align}$

Just need to prove that $\quad b^4+(b+1)^4\mod 7\neq 0$ which is quite straightforward (plug and check result for values $b=0\cdots 6$).

Answer 2

We can restate the equation as $7a^3+63a^2+273a+441=2b^4+4b^3+6b^2+4b+1$, so $7|2b^4+4b^3+6b^2+4b+1$. But you can verify none of the residue classes modulo $7$ achieve this: from $b\equiv0$ to $b\equiv6$, the remainders of $2b^4+4b^3+6b^2+4b+1$ are $1,\,3,\,6,\,1,\,6,\,3,\,1$.