# Diophantine equation: $2 a^2 + 2 b^2 = c^2 + d^2$ [duplicate]

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.

• The equations you have written define a conic in $\mathbb{P}^2$ with at least one solution. Hence there are infinitely many primitive solutions, which can be parametrised by using sterographic projection. The situation here is very similar to what happens with Pythagorean triples. – Daniel Loughran May 29 '14 at 15:23
• @DanielLoughran: nice solution; IMHO it would be useful to expand it into an answer that includes the birational map you mentioned. – Michael May 29 '14 at 16:04
• @DanielLoughran: actually this is a quadric in $\mathbb{P}^3$ (but of course the stereographic projection still works). – abx May 29 '14 at 16:13
• @abx: I was considering the "preferable solutions", given by both equations. – Daniel Loughran May 29 '14 at 16:29
• @Michael: I will add further details if the OP is interested, however I guess my point was that it should be easy to find lots more solutions, which is what the OP wanted. Given that this has 3 votes to close now however, I'm not sure this question will survive. Indeed it is probably too elementary for mathoverflow, but would be a good question for math.stackexchange. – Daniel Loughran May 29 '14 at 16:31

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.