find a parametric solutions for a special equation

Question

Let $a,b,c$ be rational.Find a rational parametric solutions for $a,b$ and $c$ so that simultaneously $$a^2-c^2=\square$$ and $$b^2-c^2=\square$$

Answer 1

Suppose $a^2-c^2=m^2$ and $b^2-c^2=n^2.$ Define $A=a/c,\ B=b/c,\ M=m/c,\ N=n/c.$ Then the equations are $A^2-1=M^2,\ B^2-1=N^2.$ We can now use the relation $$(u^2-1)^2+(2u)^2=(u^2+1)^2 \tag{1}$$ to get (possible) formulas for $A,B,M,N.$ $$A=\frac{u^2+1}{2u},\ M=\frac{u^2-1}{2u},\\ B=\frac{v^2+1}{2v},\ N=\frac{v^2-1}{2v}. \tag{2}$$ These must each be multiplied by $c$ to get the $a,b,m,n$ going with a particular $c.$ These are rational solutions, and there is a lot of freedom in the choices. Also, one could on one or the other of these divide $(1)$ by $(u^2-1)^2$ instead of $(2u)^2$ and obtain other different looking expressions, though they may really be equivalent anyway.

Answer 2

This system of equations:

$$\left\{\begin{aligned}&a^2+b^2=c^2\\&z^2+b^2=x^2\end{aligned}\right.$$

Solutions have the form:

$$b=4tkp^2s^2$$

$$a=2(t^2-k^2)p^2s^2$$

$$z=4k^2s^4-t^2p^4$$

$$c=2(t^2+k^2)p^2s^2$$

$$x=4k^2s^4+t^2p^4$$

$t,k,p,s$ - integers asked us.