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I was just wondering if the rough outline of this proof was logically rigorous.

If a and b are integers then $a^2 -4b-2$ does not equal zero.

Proof:

Suppose for the sake of contradiction that $a^2-4b-2=0$. Then we have $a^2 = 2(2b+1)$. $a^2$ is even which implies that $a$ is even. If $a$ is even then $a = 2k$ for some integer $k$. $(2k)^2 = 2(2b+1)$ $2(2k^2) = 2(2b+1)$ $2k^2 = 2b+1$ Let $w = 2k^2$ where since$ k^2 $is an integer, $w$ is even. Let $v = 2b+1$ where since $b$ is an integer, $v$ is odd. Then we have, $w = v$, which is a contradiction.

QED

Thanks for the feedback!

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    $\begingroup$ Simple. If $a^2=4b+2$, then $2$ divides $a^2$ but $4$ doesn't divide $a^2$! This is not possible. That's a short version of your proof, which is correct. $\endgroup$ – астон вілла олоф мэллбэрг Aug 2 '16 at 0:29
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    $\begingroup$ Very good. But personally I wouldn't bother introducing $v$ and $w$. After $2k^2=2b+1$ I would just say "this is a contradiction because the left hand side is even and the right hand side is odd". $\endgroup$ – David Aug 2 '16 at 0:31
  • $\begingroup$ Your proof is fine. This kind of proof can be used to prove $\sqrt 2$ is an irrational. $\endgroup$ – Zack Ni Aug 2 '16 at 0:36
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I think the best way to think about this is to work modulo 4:

$0^2 \equiv 0$ mod 4

$1^2 \equiv 1$ mod 4

$2^2 \equiv 0$ mod 4

$3^2 \equiv 1$ mod 4

Clearly $4b + 2 \equiv 2$ mod 4. So indeed it is not possible that $a^2$ = 4b + 2. Similarly you can show $a^2$ - 4b - 3 is not equal to zero either.

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