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I am trying to solve $a^3+b^5 = 7^{7^{7^7}}$.

I have proven that $7^{7^{7^7}} \equiv 19\mod31$ using Fermat-Euler Theorem. A previous answer on this site asserts that $19$ is not equivalent to the sum of a cube and fifth power $\mod31$. Is there a better way to prove that than to just list the cubic and quintic residues? I'm afraid that's all I've come up with and it's quite tedious.

Thank you.

John Smith
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I don't think it's necessarily all that tedious. Since $31$ is prime, the multiplicative group module $31$ is cyclic of order $30$, so if you can find a generator $g$ for it, the cubes are exactly the ten powers of $g^3$, and the fifth powers are exactly the six powers of $g^5$.

So all you need to do is to compute a list of the cubes, and then check that $19-g^{5n}$ is not among them for $0\le n\le 5$.

You only need to actually multiply out half of the cubes and fifth powers, because then the rest are minus the ones you've already found. (That is, $g^{15}=-1$).

hmakholm left over Monica
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  • Ahh I did use that method - clearly I don't have enough patience whoops. This was a non-calculator question though, which I why I asked if I'd missed a fancier technique. Thank you very much for the explanation. – John Smith Sep 04 '17 at 20:35
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    By brute force, $12$ and $19$ are the only numbers that cannot be written $a^3 + b^5$ modulo 31. – Gribouillis Sep 04 '17 at 20:40