Diophantine equation: $2 a^2 + 2 b^2 = c^2 + d^2$

205 Views Asked by At

I am looking for integer solutions of the following equation: $$ 2a^{2} + 2b^{2} - c^{2} - d^{2} = 0 $$ Preferentially the solutions should obey $a+b+c+d=0$.

By inspection I found the solutions: $(a,b,c,d)=(1,0,1,1)$, and $(a,b,c,d)=(1,1,2,0)$. Additional solutions can be generated by changing the signs of $a,b,c,d$, or by scaling by an integer, or by swapping $a$ and $b$, and $c$ and $d$.

As a theoretical physicist I am rarely working with diophantine equations, and hence I am wondering what other solutions exist that I have not taken into account.

1

There are 1 best solutions below

0
On

The solutions to $2(a^2+b^2)=c^2+d^2$ and $a+b=c+d$ are essentially $$c,d=4P^2,-2Q^2,a,b=2P^2\pm 2PQ-Q^2. $$ For a fuller solution see the duplicate question. Perhaps someone with higher privilege here can merge these two.