<s>[INST] 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?
 [/INST] Let  $a^2 + 1$ be odd and $a^2 + 1$ not be prime.  The proof glibly assumes such numbers are easy to find by pointing out any $2 + 5k$ will be such a number. 
Since $a^2 + 1$ is not prime nor then $a^2 + 1 = r*s$ for some odd $r, s$ and $s \le r$.
Let $x$ be the midpoint between $s$ and $r$.  (In other words $\frac {r+s}2$).  This midpoint is an integer because $r$ and $s$ are both odd.
Let $b = x-s = r-s$.  This means $r = x+b$ and $s= x-b$.  That's not surprising as $x$ is the midpoint and $b$ is the distance each is from the midpoint.
Consider the three consecutive numbers $n = x^2 -1$, $n+1 = x^2$ and $n + 2 = ^2 + 1$.
Obviously $n+1 = x^2 + 0^2$ is the sum of two squares. And so $n+2=x^2 + 1^2$.
But $n$ is as well because:
$a^2 + b^2 = (a^2 + 1) + b^2 - 1$
$= r*s + b^2 -1= (x+b)(x-b) + b^2 - 1$
$= x^2 -b^2 + b^2 -1 = x^2 - 1 = n$
So $n$ is the sum of $a^2+ b^2$.
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