Question is: If x and y are positive numbers with x >y+1 then x^2 -y^2 is not prime.
I attempted a proof by cases for this problem. using 4 cases: x=odd, y = odd x=odd, y = even x=even, y = even x=even, y= odd
for the first case (x=odd, y =odd) I said: let x = (2a+1) + 2 I added the 2 because at every instance x must greater than y by at least 2 (x>y+1). let y = 2b+1
x^2-y^2 = (2a+3)^2 - (2b+1)^2 After simplifying my final answer was 4(a^2-b^2+3a-b+2).
So in my proof I said(this isnt my whole proof just a summary) since no matter what ever a or b is the answer will always be even and greater than 2. Therefore making it not a prime number and proving the proposition.
Would this be right? Would I have to continue onto the other cases or would that be redundant? What if another case proves that it can be a prime number?
Case 2 proved that it can be a prime number x=odd, y= even let x= (2a+1)+2 let y = 2b
(2+3)^2-(2b)^2 = 4(a^2-b+3a) + 9
Since an even number + an odd number is always an odd number that makes the possibility of the result being prime which contradicts the first case. Does this mean I have to use another method to prove this?
If both $x$ and $y$ are odd numbers with $x>y+1$, then the form of $x$ and $y$ should be $x=2a+1$ and $y=2b+1$ where $a>b$. Just adding $2$ to $x$ without specifying the order of $a$ and $b$ would not be correct.
Proof by cases may not be the easiest approach for solving this problem unless you are told to do so. A direct method might be more fruitful. Just use the facts that both $x-y$ and $x+y$ are positive integers $>1$ and $x^2-y^2=(x-y)(x+y)$.