Prove by contradiction that if $n^3$ is a multiple of $3$, then $n$ is a multiple of $3$ Problem statement:
Using proof by contradiction, prove that if $n^3$ is a multiple of $3$ , then $n$ is a multiple of $3.$
Attempt 1:
Assume that there is exist $n$ which  is a multiple of $3$ such that $n^3$ is not a multiple of $3.$
Then $n = 3k $ , $n^3 = 27 k^3 $ which a multiple of $3$, which contradicts the assumption that $n^3$ is a multiple of $3.$
Attempt 2:
Assume that there exist $n^3$ which is a multiple of $3$ such that $n$ is not a multiple of $3.$
Then $n = k+ 1 $ or $n= k+2.$
Then $n^3 = k^3 + 3k^2 + 3k + 1$
or $n^3 = k^3 + 6 k^2 + 12 k + 8.$
Both cases contradict the assumption that $n^3 $ is a multiple of $3.$
Which solution is correct ? Or do both work ?

If $n$ is not a multiple of $3$, then $n$ can be $3k+1$ or $3k+2$.
If $n = 3 k + 1$ then $n^3 = 27 k^3 + 27 k^2 + 9 k + 1 = 3 (9 k^3 + 9 k^2 + 3k) + 1$
; so $n^3$ is not divisible by $3$.  If $n = 3 k + 2$ then $n^3 = 27 k^3 + 54 k^2 + 36 k + 8 = 3 (9 k^3 + 18 k^2 + 12 k + 2) + 2 $.  So, again, $n^3$ is not divisible by $3$.  Thus we have be contradiction that if $n^3$ is a multiple of $3$ then so is $n$.
