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?

  • $\begingroup$ I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$ $\endgroup$ – saulspatz Jan 24 at 16:26

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.

  • $\begingroup$ I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v \neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$? $\endgroup$ – Phi_24 Jan 24 at 17:38

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