Solutions to $a(m^2-a^2)=b(m^2-b^2)$

Question

This problem has to do with congrua. First consider the equation: $$x^2-y^2=y^2-z^2$$ Which is 3 squares that from an arithmetic sequence. The common difference is called a congruum. I'm looking to parametrize pairs of equal congrua, meaning a congruum such that there are 2 arithmetic sequences (or more). For example: $840$ $$47^2-37^2=37^2-23^2=840=41^2-29^2=29^2-1^2$$ Individual congruum can been parametrized as follows: $$x=m^2+2mn-n^2$$ $$y=m^2+n^2$$ $$z=n^2+2mn-m^2$$ And the conruum itself is: $4mn(m^2-n^2)$

For all congruum, either there exists integer $m,n$ as shown above or the congruum is the product of a square and another congruum (for which $m, n$ exist).

With that out of the way, for a pair of equal congrua, there will exist $m,n,p,q$ such that: $$mn(m^2-n^2)=pq(p^2-q^2)$$ If this can be parametrized then great! If not, then it would still be great help to parametrize solutions for this simpler equation. (Which the same equation as above except choose $m=p$) $$n(m^2-n^2)=q(m^2-q^2)$$ One solution is:$m=7,n=3,q=5$

Note: $$n(m^2-n^2)=q(m^2-q^2)\implies n(m^2-n^2)=q(m^2-q^2)=(n+q)((n+q)^2-m^2)$$

Answer 1

Your equation $$a(m^2 - a^2) = b(m^2 - b^2)\tag1$$ by factoring is equivalent to solving $$a^2+ab+b^2=m^2$$ Do the transformation $a=e-f,\;b=2f$ and it reduces to, $$e^2+3f^2=m^2$$ The complete solution to this is $$(e,\;f,\;m) = (u^2-3v^2,\;2uv,\; u^2+3v^2)$$ For your more general equation $$pq(p^2-q^2)=rs(r^2-s^2)\tag2$$ This is equivalent to $$(p+qi)^4+(r-si)^4=(p-qi)^4+(r+si)^4$$ with the imaginary unit $i=\sqrt{-1}$. This was considered by Euler et al and can be solved by an elliptic curve, hence there are infinitely many parameterizations. One small example by L. Lander is, $$(p,\;q,\;r,\;s) = (v^5-2v,\;v^5+v,\;-2v^4+1,\;v^4+1)$$ From this initial rational point, one can find an infinite more.