Positive integers $(a, b, c)$ are a primitive Pythagorean triple

Question

Show that if $a = m^2 - n^2$ , $b = 2mn$, $c = m^2 + n^2$ , where $m$, $n$ are relatively prime, not both odd, and $m>n$, then $(a, b, c)$ is a primitive Pythagorean triple.

This is part one of a proof I am required to do.

I know that if $m$ and $n$ are not both odd, then they can be written as $2k+1$ and $2l$, respectively, or as $2k$ and $2l$. I plugged in the given values for a, b, and c into the equation $a^2 + b^2 = c^2$ and got $m^4 + 2m^2n^2 + n^4$, but this is as far as I can get, however. I know that to show $a,b,c$ are primitive I need to show their GCD is $1$, but I don't know how to do this. Can someone show me where to start?

Answer 1

Line up your ducks. And then shoot them.

Does

$(m^2 - n^2)^2 + (2mn)^2 {? \over=} (m^2+n^2)^2$

$m^4 - 2m^2n^2 + n^4 + 4m^2n^2 {? \over=} m^4 + 2m^2n^2 + n^4$

$m^4 + 2m^2n^2 {? \over=} m^4 + 2m^2n^2 + n^4$?

The answer is... yes, it does.

So $m^2-n^2, 2mn, m^2 + n^2$ are a pythogorean triple.

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But are the a primative triplet? That is:

And are $m^2 - n^2$ and $2mn$ relatively prime if $m,n$ are and they are not both odd?

If $p$ is a prime divisor that divides $2mn$ then either

1. $p|2$ so $p=2$.

But $m,n$ are relatively prime so they are not both even and they are not both odd so $m^2 -n^2$ is odd and so $p\not \mid m^2 - n^2$.

2. $p|m$

But $m,n$ are relatively prime $p\not \mid n$. So $p|m^2$ but not $n^2$ so $p \not \mid m^2 -n^2$.

3. $p|n$

Same argument. $p\not \mid m$ so $p|n^2$ but not $m^2$ and therefor $p\not \mid m^2 - n^2$.

so no prime factor of $2mn$ is a facctor of $m^2 - n^2$ so $m^2-n^2$ and $2mn$ are relatively prime.

So $m^2-n^2, 2mn, m^2+n^2$ is a primitive pythagorean triplet.

Answer 2

Identifying primitives is easier if we use an alternative to Euclid's formula which is the result of replacing $F(m,k)$ with $F(2m-1+k,k)$ Thw formula generates only the subset of triples where $GCD(A,B,C)$ is an odd square. The formula works for any pair of natural numbers with no trivial triples and with fewer "multiples" than Euclid's formula.

$$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$

1. If $n=1$, then $(2n-1)^2=1$ and all triples are primitive because $C-B=1$.

2. If $(2n-1)$ is prime, the following generates $\big((2n-1)-1\big)$ primitives, then skips a "multiple". This results in only primitives being generated for any k-value.

\begin{align*} &A=(2n-1)^2+&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad\\ &B=&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad+2\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)^2\\ &C=(2n-1)^2+&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad+2\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)^2 \end{align*}

3. If $(2n-1)$ is composite, there will be a primitive whenever $GCD\big((2n-1),k\big)=1$.

No further primitive tests are needed. If you want to convert back to Euclid's formula, then let $$F(m,k)=F\bigg(\frac{n+1-k}{2},k\bigg)$$