lebesgue's identity

In the Lebesgue's identity $$ a^2+b^2+c^2=d^2$$ where: $$a=m^2+n^2-p^2-q^2$$ $$b=2(mp+nq) $$ $$c=2(mq-np) $$ $$d=m^2+n^2+p^2+q^2 $$how can we write $(m,n,p,q)$ as a function of the integers $(a,b,c,d)$? I have been trying and failing. Please help.

Solution 1:

Lebesgue's Identity is complete only if you add a scaling factor t. Hence,

$$\begin{align} a &= (m^2+n^2-p^2-q^2)t\\ b &= 2(mp+nq)t\\ c &= 2(mq-np)t\\ d &= (m^2+n^2+p^2+q^2)t \end{align}\tag{1}$$

As a function of the integers $a,b,c,d$, then the variables $m,n,p,q,t$ are simply,

$$\begin{align} m &= \frac{-b-c}{a-d}\\ n &= \frac{-b+c}{a-d}\\ p &= 1\\ q &= 1\\ t &= \tfrac{1}{4}(-a+d) \end{align}$$

If you substitute these into the eqns in $(1)$, for example,

$$(m^2+n^2-p^2-q^2)t-a = 0$$

then factor it, you get this result, hence is true if indeed $a^2+b^2+c^2 = d^2$. (Similarly for the 2nd, 3rd, 4th eqns.)

For the smallest solution,

$$1^2+2^2+2^2 = 3^2$$

we get $m,n,p,q,t = 2, 0, 1, 1,\tfrac{1}{2}$. And so on.