2

Is this possible to provide full parametrization for the following Diophantine equation of order 2:

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

as we can give all the solutions for:

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

as

$(a=m^2-n^2$, $b=2mn$, $c=m^2+n^2)$

It is stated that if you can find a non-trivial solution then you can find all solutions of Diophantine equations of order 2.

A non-trivial solution would be $(2,1,3,1)$.

heartwork
  • 129
  • Definitely these are not all solutions. For example for some given (a,b) we may have multiple pairs (c,d). You did not understand my question. – heartwork Sep 04 '24 at 08:50
  • See also this old post though the answer below is much simpler, and easier seen to be complete via the Brahmagupta-Fibonacci-Euler identity. – Tito Piezas III Mar 01 '25 at 07:15

1 Answers1

2

Yes, this has a complete solution like $a^2+b^2=c^2$. Given,

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

we first do the minor change of variables $a=\dfrac{e+f}2\,$ and $\,b=\dfrac{e-f}2$ to get,

$$c^2+d^2=e^2+f^2$$

We insert $n$ to make it more general,

$$c^2+nd^2=e^2+nf^2$$

This symmetric form is well-known to have a complete solution by Euler and earlier by Brahmagupta. It is given by,

$$(c,d,e,f) = (pr + nqs,\, ps - qr,\, pr - nqs,\, ps + qr)$$

for four arbitrary variables $(p,q,r,s)$ such that,

$$c^2+nd^2=e^2+nf^2=(p^2+nq^2)(r^2+ns^2)$$

where we just set $n=1$ to focus on the original equation.

Tito Piezas III
  • 60,745