Number of Pythagorean Triples

Question

I am trying to solve an exercise from the book "Theory of Numbers" by B.M.Stewart. The exercise is the following one:

Let $T=2^ap_1^{a_1}p_2^{a_2} \dots p_n^{a_n}$, where $a \ge0, n\ge0, 2<p_1<p_2<\dots p_n, p_j$ odd prime numbers $ \forall j=1 \dots n,a_j \ge1$ and let $S(T)$ indicate the number of primitive Pythagorean triplets of side $T$. Show that $S(T) = 2^{n-1}$ if $a=0$.

The primitive Pythagorean triplets are the solutions of $x^2+y^2=z^2$ where $\gcd(x,y,z)=1$ and they are given by $$\cases {x=2uv \\y=u^2-v^2\\z=u^2+v^2\\u>v>0\\ \gcd(u,v)=1\\ u \not\equiv v \pmod2} $$

If $a=0$ then $T=p_1^{a_1}p_2^{a_2} \dots p_n^{a_n}$, so $T \ne 2uv$. Then $T=y$ or $T=z$.

If $T=y$, then $$T=u^2-v^2=(u+v)(u-v)=p_1^{a_1}p_2^{a_2} \dots p_n^{a_n}.$$ So for $(u+v)$ I have $2^n$ possibilities because I can count the number of prime factors which is $\tau(r)$ with $r=p_1p_2 \dots p_n$.

From there I don't know how to proceed. Have you any idea?

Answer 1

You're awfully close, if you accept my comment that only the $T=y$ case needs to be considered. It follows from $\gcd(u,v)=1, u \not\equiv v \pmod2$ that $\gcd(u+v,u-v)=1.$ There are $2^n$ ways to split $T$ up into a pair of relative prime factors, and the larger one must be $u+v,$ so you just solve two linear equations for $u$ and $v$.

EDIT Suppose we had $T=3^2\cdot5\cdot7.$ We could split this into two the factors $9$ and $35$. Then we would have to solve $$\begin{align}u+v &=35\\u-v&=9\end{align}$$ so $u=22,v=13.$

We have to divide the number of solutions by $2$ because we could always have to make $u+v$ the larger factor, so half the choices are inadmissible.

Answer 2

I'm not sure this helps because I don't understand the function completely. The formula I've alway seen for generating triples is: $$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2$$ and it is useful but it generates extraneous and trivial triples if $m\le n$; it also generates doubles and odd-or-even square multiples of primitive triples. We can generate triples for all $m,n\in\mathbb{N}$ where $GCD(A,B,C)$ is an odd square (which include all primitives) and get distinct $sets$ of triples by replacing $(m,n)\text{ with }((2m-1+n),n).$ Expanded, this becomes:

$$A=(2m-1)^2+2(2m-1)n\quad B=2(2m-1)n+2n^2\quad C=(2m-1)^2+2(2m-1)n+2n^2$$ and the result is as shown in the sample below.

$$\begin{array}{c|c|c|c|c|} \text{$Set_n$}& \text{$Triple_1$} & \text{$Triple_2$} & \text{$Triple_3$} & \text{$Triple_4$}\\ \hline \text{$Set_1$} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline \text{$Set_2$} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline \text{$Set_3$} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline \text{$Set_{25}$} &2499,100,2501 &2597,204,2605 &2695,312,2713 &2793,424,2825\\ \hline \end{array}$$

In all cases, $m$ is the set number and $n$ is the $n^{th}$ member of the set.

We can see that $f(2,3)=(27,36,45)$ is $9\text{ times }(3,4,5)$ and this is because $(2m-1)=(2*2-1)=3$, and $GCD((2m-1),n)=3.$ Whenever $GCD((2m-1),n)=1$, a primitive will be generated; otherwise, an odd square multiple will be generated. Only in $Set_1 (m=1)$ or (rotated $90^\circ$), in the first member of each set $(k=1)$, are all triples primitive and these can be generated by $$\text{For }m=1:\quad A=2k+1\qquad B=2k^2+2k\qquad C=2k^2+2k+1$$

$$\text{For }k=1:\quad A=4m^2-1\qquad B=4m\qquad C=4m^2+1$$

Note that prime values of $C$ can appear in all sets but prime values of $A$ can only appear in $Set_1(m=1)$ because $$A=(2m-1)^2+2(2m-1)n=(2m-1)(2m-1+2n)$$ so $m>1$ results in a composite number $A$.

If you need help finding triples for given sides see this.

I hope this helps in your search for primitives and primes.