discrete mathematics – Prove that there are not 2 integers \$ a \$ and \$ b \$ such that \$ a ^ 2-b ^ 2 = n \$

so I have this question:
Let n be an even integer.

(a) We assume that $$n / 2$$ is odd. Prove that there are not two integers $$a$$
and $$b$$ such as $$a ^ 2-b ^ 2 = n$$

(b) We assume that $$n / 2$$ is same. Prove that there are two integers a and b
such as $$a ^ 2-b ^ 2 = n$$

we start by saying that a and b must both be odd or even so that n is even if a and b are then equal $$a ^ 2$$ and $$b ^ 2$$ are also and $$a ^ 2 = 4k ^ 2$$ and $$b = 4L ^ 2$$ then we get that $$4 (k ^ 2-L ^ 2) = n$$ => $$2 (k ^ 2-L ^ 2)$$=$$n / 2$$ which is a contradiction because even does not mean odd and we repeat the process with a and b are weird we get the same result meaning that they do not exist

as for B. we do the same thing as A but this time, it will end with the same = even if a and b exist

it's the right solution?