Pythagorean triples, proof clarification

Question

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 $4n^2 + 4n$.

Answer 1

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 $b=2n^2+2n$.

Answer 2

That proof is showing you the author's reasoning - how they discovered that choosing $$ b = 2n^2 + 2n $$ would do the job.

They could have written the proof as a straightforward implication starting with that definition of $b$. But without the preliminary analysis you could follow the algebra but reasonably ask "where did that come from"?

Answer 3

The problem is with the word "then" just before your long equality, and the text after. It should be replaced with something like this:

If such an integer $b$ exists, then $a, b,$ and $b+1$ must satisfy the following equality.

$$equation here$$

which is equivalent to $2b = 4n^2 + 4n$. Picking $b = 2n^2 + 2n$, we see that there is indeed such a number, hence...

NB: seeing Ethan's answer, I realize that you're quoting from someone else's proof that confused you. I thought you were quoting your own proof, which had been marked correct, but then gave you pause as you looked at it.

Answer 4

We begin with Euclid's formula which is probably the most accepted one for generating Pythagorean triples. $$A=m^2-k^2 \qquad B=2mk \qquad C=m^2+k^2$$ If we replace $\,m\,$ with $\,(2n-1+k),\,$ we get $$A=(2n-1+k)^2-k^2 \quad B=2(2n-1+k)k \quad C=(2n-1+k)^2+k^2$$ which resolves to \begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+&2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+&2k^2 \end{align*} and generates the subset of triples (below) where $\,C-B=(2n-1)^2.\quad$ By inspection of both the formula and the table, we can see that $\,B\,$ is always a multiple of $\,4\,$ (even) and that $\,n=1\implies C-B=1.$

\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 \\ \hline \end{array}