This equation has no solutions in $\mathbb{Z}/43\mathbb{Z}$ so it has no solutions in integers.

can you pls illustrate??

The possible residues of $x^6\mod43$ and $y^7\mod43$ are 1,21,41,11,16 and 1,42,37 respectively. One manually checks $2x^6+y^7$ mod 43 for all combinations of one of the first and second group to verify.

AwesomeToad wrote: The possible residues of $x^6\mod43$ and $y^7\mod43$ are 1,21,41,11,16 and 1,42,37 respectively. How do you know it? Do you have to try all 43 posibilities or is there more elegant way?

Radar wrote: AwesomeToad wrote: The possible residues of $x^6\mod43$ and $y^7\mod43$ are 1,21,41,11,16 and 1,42,37 respectively. How do you know it? Do you have to try all 43 posibilities or is there more elegant way? They repeat in a cycle after a few terms. Try it yourself.

I'll rephrase Radar's question. How can you prove that the cycle exists as we notice it? "Noticing" is different from "knowing and proving". What's the proof behind it?

ahaanomegas wrote: I'll rephrase Radar's question. How can you prove that the cycle exists as we notice it? "Noticing" is different from "knowing and proving". What's the proof behind it? Its real easy to prove for lower mods because the forms(described right below) are limited. All integers come in the form of $3k, 3k+1, 3k-1$ for some integer $k$. When we cube them, we have: $(3k)^3 \equiv 0 \pmod 3$ $(3k+1)^3 \equiv 1 \pmod 3$ $(3k-1)^3 \equiv 2 \pmod 3$ So all possibilities are only $0,1,2$ When you have higher mods, its considerably more difficult and I don't know how to prove those.

When you have a higher modulus its trivial by using primitive roots because there only so many values $w^{kn}$ can take when $n|(p-1)$ and $w$ is a primitive root while $k$ is a varying number. There is a much less overkill proof, but right now I don't have the time to write it out.