Let $m$ and $n$ be integers. Then $n^3 – 2m – 2 \ne 0$.

Question

I'm supposed to prove this using proof by contradiction, and I'm having a lot of trouble with it.

I understand that I'm supposed to assume the premise is true and the conclusion false, which would mean the new statement would be "Let m and n be integers. Then $n^3 - 2m - 2 = 0$. I've been trying to switch up the equation by adding $2m+2$ to the other side of the equation, but I really can't figure out what to do next. A hint that was given to me by my Professor was to try and contradict the premise by showing that $n, m$ or both are not integers.

Answer 1

Note that

$$n^3 – 2m – 2 = 0\iff n^3=2(m+1)$$

thus $n^3$ must be even let $n=2k$

$$\iff n^3=2(m+1)\iff 8k^3=2(m+1)\iff 4k^3=m+1$$

$$\iff 4k^3-1=m$$

we don't find any contradiction and thus we can find infinitely many solutions: $$(n=2, m=3), (n=4, m=31), etc.$$