Pythagorean triples, proof clarification Question
Prove for every odd integer $a \geq 3$ that there exists an even integer $b$ such that $(a, b, b + 1)$ is a
Pythagorean triple.
Proof
Let $a \geq 3$ be an odd integer. Then $a = 2n + 1$ for some positive integer $n$. We seek
a positive even integer $b$ such that $(a, b, b + 1)$ is a Pythagorean triple. Then
$a^2 + b^2 = 4n^2 + 4n + 1 + b^2 = b^2 + 2b + 1.$
Therefore,
$2b = 4n^2+4n$ and so $b = 2n^2+2n.$
Letting $b = 2n^2+2n$ , we see that $a^2+b^2 = (b+1)^2$
and so $(a, b, b + 1)$ is a Pythagorean triple.
My question is how we could we write $4n^2 + 4n + 1 + b^2 = b^2 + 2b + 1$ before we say that $2b = 4n^2+4n$? The way it was written makes it seem like $4n^2 + 4n$ naturally simplifies to $2b$, but I don't see how it does before $2b$ is explicitly defined to equal $4n^2 + 4n$.

This is actually a pretty common technique of proof. We're essentially working backwards from the conclusion: if there was a $b$ satisfying $a^2+b^2=(b+1)^2$, then necessarily we would have $4n^2+4n+1+b^2=b^2+2b+1$ by expanding the two expressions. The $b^2$ terms cancel and we would find that $4n^2+4n=2b$ and therefore $2n^2+2n=b$. To prove your claim, we need to supply the correct number $b$, but this calculation shows us that there is only one possibility, namely take $b=2n^2+2n$.
