<s>[INST] Prove that $ a^2-4b \neq2$ if $ a,b \in \mathbb{ Z}$ My solution :
We suppose that is true. Then by contradiction:
$a^2-4b-2=0$
$a^2=4b+2$
$a=2(b+1/2) ^{0.5}$
then $(b+1/2)$ is fraction and rooted by $0.5$ so the square root of any fraction $+$ any-Integer will give fraction so then $a$ must be fraction, not an integer. Contradiction.
Then $a^2 -4b \neq 2$?
Is that a good proof?
 [/INST] Suppose
$a^2=4b+2
$.
$a$ must be of the form
$2c$ or $2c+1$
(i.e., even or odd).
If $a=2c$,
then
$2 = (2c)^2-4b
= 4c^2-4b
= 4(c^2-b)
$.
But 4 divides the right side but not the left,
so this is impossible.
If $a=2c+1$,
then
$2 = (2c+1)^2-4b
= 4c^2+4c+1-4b
= 4(c^2+c-b)+1
$.
But the left side is even
and the right side is odd,
so this is also impossible.
Note that this works,
with minor changes,
for
$a^2 = 4b+3
$.
</s>