Need help to understand the following number theory proof Prove that there exist infinitely many integers n such that $n, n + 1, n + 2$ are each the sum of the squares of two integers:
Solution: First solution: Let a be an even integer such that $a^2 + 1$ is not prime. (For example, choose a ≡ 2 (mod 5), so that $a^2 + 1$ is divisible by 5.) Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$, and setting x = (r + s)/2, b = (r − s)/2. Finally, put n = $x^2 − 1$, so that $n = a^2 + b^2$, $n + 1 = x^2$, n + 2 = $x^2 + 1$
I got lost in the following part:
Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$. What is rs?. Can someone expand this proof, so it looks clearer?

Just follow the instructions:


*

*Pick some even $a$ such that $a^2+1$ is not a prime, for instance some $a\equiv 2\pmod{5}$.
Fine, let us pick $a=22$;

*Write $a^2+1$ as $r\cdot s$ with $r\geq s >1$.
Fine, $22^2+1 = 97\cdot 5$;

*Write $a^2+1$ as a difference of squares through the previous decomposition.
Fine, $22^2+1 = (51+46)(51-46) = 51^2-46^2$.


This gives that $51^2-1$ is the sum of two squares, $22^2+46^2$, so $51^2-1,51^2$ and $51^2+1^2$ are three consecutive numbers belonging to $\square+\square$. Is it a bit clearer now?
