# Number of Pythagorean Triples

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 $$a \ge0, n\ge0, 2 odd prime numbers $$\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?

• 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 $T=z.$ – 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.
• 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 $u+v$? – Phi_24 Jan 24 at 17:38