My tittle is: Don't change my tittle.

Althought it's a cuasi-particular Pell's equation or this link I'll do my "own proof"

a) I'm going to prove that $\forall p > 2$ with p an odd natural number , there exists at least 4 solutions for the diophantine equation $p = x^2 - y^2 = (x - y)(x + y)$ Take $\frac{p + 1}{2} = x$ and $y = x - 1 \ge 1$ , then \begin{cases} {x - y = 1 \\ y + x = p} \end{cases} \Rightarrow p = x^2 - y^2 = (x -y)(x + y) has at least one solution. Changing $x$ to $- x$ , and $y$ to $-y$ we get at least 4 solutions.

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b)Let's suppose $2 = x^2 - y^2 = (x - y)(x +y)$ has at least one solution with $x,y \in \mathbb{Z} \Rightarrow$ we'll have 4 posibilitties which lead us to contradictions.

1.- Let's suppose \begin{cases}{x - y = 1 \\ x + y = 2}\end{cases} \Rightarrow \begin{cases}{ x= y + 1 \\ 2y + 1 = 2}\end{cases} \Rightarrow y = \frac{1}{2} . This is a contradiction, $y$ must belong to integers numbers set.

2.- Let's suppose $x - y = - 1$ ...

3.- Let's suppose $x - y = 2$ ....

4.- Let's suppose $x - y = -2$ ...

Part a: Any odd number $2m+1$ can be written as $2m+1=(m+1)^2-m^2$

Part b: Assume $x>y>0$ without loss of generality. Then $x^2-y^2\geq (y+1)^2-y^2=2y+1\geq 3$