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?