How do I use proof by contradiction to prove that for all prime numbers $x$, $y$, and $z$, $x^2 + y^2 \neq z^2$?

Question

Original implication: For all prime numbers $x$, $y$, and $z$, $x^2 + y^2 \neq z^2$.

I'm not certain if I'm understanding the process of proof by contradiction correctly. What I am understanding so far is that I must first make the initial statement a contrapositive. Which would be:

$x^2 + y^2 = z^2 \Rightarrow$ some $x$, $y$, $z$ belonging to $\mathbb{Z}$ (integers), not($P(x, y, z)$), where $P$ = prime numbers.

I would greatly appreciate help in figuring out the rest of the steps to prove by contradiction.

score 5 · Answer 1 · answered Mar 04 '15 at 04:17

5

Hint: Suppose that $x^2+y^2=z^2$ for some prime $x,y,z$. Not all of $x,y,z$ can be odd (exercise), so at least one must be even, i.e., 2. Now there are two cases: either $x=2$ or $z=2$. The latter case is easy to rule out. For the former, obtain a contradiction by considering $4 = z^2-y^2 = (z-y)(z+y)$ and deducing the possible values of $y,z$.

answered Mar 04 '15 at 04:17

Yuval Filmus

57,157

I was trying to figure out which two square numbers subtract to make 4, but I don't think such numbers exist... Are there? – pudi Mar 04 '15 at 04:30
Try to use the identity $4 = (z-y)(z+y)$. – Yuval Filmus Mar 04 '15 at 04:30

How do I use proof by contradiction to prove that for all prime numbers $x$, $y$, and $z$, $x^2 + y^2 \neq z^2$?

1 Answers1