Factorize ANY pythagorean quadruple using 4 variables?

Question

Can ANY pythagorean quadruple be generated using 4 integer variables $(m,n,p,q)$?

I've read up on Mordell's method, which uses a 3-variable system of generating pythagorean quadruples, and it is widely documented that this doesn't generate them all. However there is also another method which introduces a 4th parameter, q. What I've read suggests that all pythagorean quadruples can be generated if you include a 4th parameter - but is it the case that even this 4-parameter method doesn't generate them all?

From wikipedia the 4-variable equations are given as:

$a^2 + b^2 + c^2 = d^2$

where

$a = m^2 + n^2 - p^2 - q^2$

$b = 2(mq + np)$

$c = 2(nq - mp)$

$d = m^2 + n^2 + p^2 + q^2$

where $a$ is odd.

Consider the pythagorean quadruple $(a,b,c,d) = (3171,172,1620,3565)$.

Using (4) - (1), I infer $2(p^2 + q^2) = d - a = 3565 - 3171 = 394$

Therefore $p^2+q^2 = 197$ and the only possible values for p and q are $(p,q)=(28,195)$.

But then by expanding (2) and (3), we have either

$56m + 390n = 1620$

$56n - 390m = 172$

OR

$56m + 390n = 172$

$56n - 390m = 1620$

whichever way round I slice (2) and (3) there seems to be no integer solutions for $m$ and $n$.

So is it either the case that not all pythagorean quadruples have integer solutions of $m,n,p,q$ or have I made an error somewhere that I am not seeing?

Answer 1

You have solve $p^2+q^2=197^2$ where you should have solved $p^2+q^2=197$.

This leads to $m=2$, $n=58$, $p=1$, $q=14$ which gives:

$$a^2=2^2+58^2-1^2-14^2=3171$$

$$b=2(2\cdot14+58\cdot1)=172$$

$$c=2(58\cdot14-2\cdot1)=1620$$

$$d^2=2^2+58^2+1^2+14^2=3565$$