If a and b are two positive integers, prove $a^2-4b \neq 2$ I know that there are answers for this question, but I didn't come across the following proof, and I was wondering if it was correct:
I re-arranged the equation: $a^2 \neq 2 \pmod 4$
Case 1: when a is even:
let $a = 2k$ where $k \in Z$
$(2k)^2 \neq 2\pmod4$
$4k^2 \neq 2\pmod4$
$4(k^2) \neq 2\pmod4$
the statement is true because LHS does not have a remainder 2.
Caes 2: when a is odd
let $a = 2k+1$ where $k \in Z$
$(2k+1)^2 \neq 2\pmod4 $
$4k^2 + 4k + 1 \neq 2\pmod4$
$4(k^2 + k) + 1 \neq 2\pmod4$
the LHS has a remainder of 1 when divided by 4 and the RHS has a remainder of 2 which does not equal $\therefore$ the statement is true
 A: I agree with Parcly Taxel's comment that it basically looks fine.
My only suggestion is to not use what you're trying to prove in each part as this makes it appear you've already shown, and thus don't need to go on, or you're assuming initially what you're trying to prove. Thus, for the first part of $a = 2k$, I would instead do something like $(2k)^2 = 4k^2 \equiv 0 \not\equiv 2 \pmod 4$.
A: Your method looks okay.  Or you could say $a\equiv0, 1, 2, $ or $3\pmod4$ so $a^2\equiv0, 1, 0, $ or $1\pmod4,$ not $2$.  
A: Our OP's proof looks fine to me, however:
We don't really need to consider the even and odd cases separately.
If 
$a^2 - 4b = 2 \tag 1$
for 
$a, b \in \Bbb N, \tag 2$
we may write
$a^2 = 4b + 2 = 2(2b + 1) \Longrightarrow 2 \mid a; \tag 3$
$2 \mid a \Longrightarrow 4 \mid a^2 \Longrightarrow 4 \mid a^2 - 4b \Longrightarrow 4 \mid 2; \tag 4$
however,
$4 \mid 2 \tag 5$
is impossible; hence, no such $a, b \in \Bbb N$ exist.  $OE\Delta.$
A: Here is an alternate take:
$a^2-4b = 2$ implies $a^2=2+4b=2(1+2b)$ and so $2v_2(a)=1$, which is impossible.
Here, $v_2(n)$ is the exponent of $2$ in the prime factorization of $n$.
