Prove the following contradiction.

If a and b are Integers then a^ 2 −4b ≠2

Proposition , then a^ 2 −4b ≠2

This is what I've done

Assume

a^ 2 −4b = 2

a^2 = 2 + 4b

a^2 = 2(2b+1)

a^2 is even because if we sub anything into b it has an even answer.

because a^2 is even it follows that a is even so a = 2k for some integer k.

now put 2k back into original assumption

(2k)^2-4b=2,

so 4k^2-4b=2 now lets devide by 2

we get 2k^2-2b=1

∴ 1 = 2(k^2-b)

because c^2-b ∈ Z it follows that 1 is even ..

so it is incorrect

we assumed the proposition false but the proposition is true.