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Prove that if $a$, $b$, and $c$ are all positive integers such that $a^2$ + $b^2$ = $c^2$ then $a$ or $b$ must be even.

One way I was taught to handle implications with an or statement was to prove that if the first part is false, then the second part must be true. What I mean by this is that I can prove this implication to be correct if I can show that if $a$ is odd, then $b$ must be even. My problem is that I don't have much to work with. If $a$ is odd, then I could split the proof into 2 cases, one where $c$ is odd and the other where $c$ is even. I couldn't seem to get anywhere with that though.

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    $\begingroup$ Have you tried assuming that both of them are odd and seeing if it leads to a contradiction? (I'm not sure it will work) $\endgroup$ – Lundborg Feb 27 '16 at 7:21
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    $\begingroup$ The "proof verification" tag means "Here's my proof. Is it right?" Since you don't have a proof, that doesn't apply. $\endgroup$ – Thomas Andrews Feb 27 '16 at 7:26
  • $\begingroup$ My fault, sorry. $\endgroup$ – Chris Feb 27 '16 at 7:28
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Suppose they are both odd. Then we get, for some integral $m,n$, $4m^2 + 4m + 4n^2 + 4n + 2 = c^2$. $c$ must thus be even, and hence it can be expressed as $c=2b$ for integral $b$. So we get

$4m^2 + 4m + 4n^2 + 4n + 2 = 4b^2$ for some $b$. Dividing both sides by four, we see $b^2$ is not integral, which implies $b$ is not integral. This is a contradiction.

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  • $\begingroup$ Alternatively, $2=4b^2-4m^2-4m-4n^2-4n$ shows that $4$ divides $2$, a contradiction. This is the same proof as user218931's, but without using modular arithmetic notation/terminology explicitly. $\endgroup$ – Jeppe Stig Nielsen Feb 27 '16 at 8:05
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If $a=2n+1$ and $b=2m+1$ are odd, then $$ a^2 = 4n^2+4n+1 = 4n(n+1)+1 \equiv 1 \mod 8 $$ because either $n$ or $n+1$ must be even. Similarly, $b^2\equiv 1\mod 8$. Now \begin{align*} a^2 + b^2 = 4n^2+4n+1 + 4m^2+4m+1 \equiv 2 \mod 8. \end{align*} But $2$ is not a square modulo $8$. Hence, there exists no $c$ such that $a^2+b^2=c^2$.

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The general parametrized solution to this diophantine equation is: $a = 2pq, b = p^2-q^2, c = p^2+q^2$. This shows that $a$ is even.

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    $\begingroup$ Showing that this parametrization is the full solution is harder than solving the problem directly, I would say. $\endgroup$ – Jeppe Stig Nielsen Feb 27 '16 at 8:00
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We can understand that what we want to prove is that

$ (odd)^2 + (odd)^2 $ will not equal to $(even)^2$

(Sum of odd numbers will only be even)

Perfect odd squares are in the form

$4n + 1$

Therefore the above expression (sum of two odd square numbers) comes down to

$ 4n + 4m + 2$

It can be proven that this will never be an even pefect square which is always a multiple of 4.

If it is, then

$(4n + 4m + 2)/4 $

will be an integer, but it is not, as it becomes

$ m + n + 0.5$

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