Proving that an equation has no solution in the set of integers.

Question

I want to show that $m^3+14n^3 = 12$ has no solution in the set of integers.

Could anyone provide any insight on how to do this? Thanks.

Answer 1

$m^3+14n^3\equiv m^3\equiv 12\equiv 5\pmod{\! 7}$, which is impossible.

$x^3\equiv \{0,1,6\}\pmod{\! 7}$ for any integer $x$, which you can see by cubing $0,\pm 1,\pm 2,\pm 3$ mod $7$.

We say $0,1,6$ are cubic residues mod $7$ and $2,3,4,5$ are cubic non-residues mod $7$.

Or you can get a contradiction using Fermat's little theorem: square both sides.

$\Rightarrow\, m^6\equiv 25\equiv 4\pmod{\! 7}$, impossible.

Answer 2

We describe an approach that is less efficient than the one of user26486, in order to introduce another theme.

Suppose that $(m,n)$ is a solution. Then $m^3$ must be even, and therefore $m$ must be even. Let $m=2s$ for some integer $s$. Then $8s^3+14n^3=12$, and therefore $4s^3+7n^3=6$. It follows that $n$ is even, say $n=2t$. We get $4s^3+8t^3=6$, which is impossible since the left-hand side is divisible by $4$, but the right-hand side is not.