I'm having issues coming up with a contrapositive proof for the following question.
As far as I know, a proof by contraposition is based on the following :
$\overline Q \to \overline P \equiv P \to Q$
or that's where I'm mistaken?
The question is:
X,Y,Z are natural numbers. if $X^3+Y^3=Z^3$ , then at least one of them is divisible by 3. Provide a proof by contrapositive.
I've tried to substitude X,Y,Z with (3A-1),(3B-1),(3C-1) , essentially trying to get to a point where:
$\overline Q$ : none is divisble by 3 (3N-1).
$\overline P$ : $X^3+Y^3\neq Z^3$
$\overline Q$ -> $\overline P$ $\equiv$ P -> Q
I tried to simplify the equation but it doesn't look like I got anywhere, where am I being wrong?
Thanks for helping!
Following the comments, if we have $X, Y$ and $Z$ none of which are divisible by 3 and such that $X^3 + Y^3 = Z^3$, we may assume that $X,Y \equiv 1$ (mod 3) and $Z \equiv 2$ (mod 3). So, we may assume that there are $a$ $b$ and $c$ such that $X = 3a+1$, $Y = 3b+1$ and $Z = 3c+2$. By applying the binomial expansion to $X^3$ and $Y^3$, it turns out that $X^3 \equiv 1$ (mod 9) and $Y^3 \equiv 1$ (mod 9), so that $X^3 + Y^3 \equiv 2$ (mod 9). But if you apply the binomial expansion to $Z^3 = (3c+2)^3$, it turns out that $Z^3 \equiv 8$ (mod 9). Thus, the equality $X^3 + Y^3 = Z^3$ cannot hold.