Solutions to the diophantine equation: $2a^2 + 2b^2- c^2- d^2 = 0$

Question

As suggested on Mathoverflow (https://mathoverflow.net/questions/168536/solutions-to-the-diophantine-equation-2a2-2b2-c2-d2-0) I am transfering this question to math-stackexchange:

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

By inspection I found the following solutions: $(a,b,c,d)=(1,0,1,1)$ , $(a,b,c,d)=(0,1,1,1)$, $(a,b,c,d)=(1,1,-2,0)$ and $(a,b,c,d)=(1,1,0,-2)$.

Additional solutions can be generated by swapping the sign of $a,b,c,d$ or by scaling the solutions that I have given by an integer.

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

I am looking forward to your responses.

Answer 1

$$a=2-2p^2-q^2-r^2+2p(q+r)$$ $$b=2(-2p+q+r)$$ $$c=-2-2p^2+4pq-q^2-2qr+r^2$$ $$d=-2-2p^2+q^2+4pr-2qr-r^2$$

Edited to add: If you also require $a+b+c+d=0$, then

$$a=p^2-2pq+q^2+2pr-2qr-2r^2$$ $$b=p^2+q^2+2qr-2r^2-2p(q+r)$$ $$c=4r^2$$ $$d=-2(p-q)^2$$

Answer 2

This should have been a comment under WillO's answer, but didn't fit.

About "conics" and such. Let's start with the most familiar "conic": a circle in $\mathbb R^2$ given by $x^2+y^2=1$, or equivalently a circle in $P^2$ given by $x^2+y^2=z^2$. Let's answer a similar question: what are all the integer triples that solve the Pythagorian equation $x^2+y^2=z^2$ (or, eqivalently, rational pair that solve $x^2+y^2=1$)?

The solution boils down to constructing a "stereographic projection" between the circle and the line. This projection gives an "almost 1:1" correspondence between rational points on the circle and rational points on a line. Here's a video that explains how Pythagorian triples are found using stereographic projection. The Geometry of Euclid's formula part of Wikipedia article explains the same thing.

The technique is extendible from the circle given by the equation $x^2+y^2=z^2$ to any "conic section" given by any quadratic equation, or in your case a system of a quadratic and linear equation. What you need to do is the following:

1) Pick a single point on the curve given by your equations which would act like a pivot for mapping your curve onto a line;

2) Pick a line in the hyperplane given by your linear equation $a+b+c+d=0$;

3) Draw a generic line through the pivot point that would intersect your curve as well as the chosen line and figure out the formula that maps between the points on the curve and the points on the line;

4) Since rational points on the line are in "almost" 1:1 correspondence with the rational points on the curve the mapping from the line to the curve would give you the formula for the integer points on the curve when you substitute the line parameter (say, "$t$") with the corresponding fraction (say, $t=m/n$).

BTW, this technique (could also be found under other names such "birational isomorphsim") is useful in field other than number theory. For example, when you are solving an indefinite integral of an expression with trigonometric functions $\sin(x)$ and $\cos(x)$ you usually make well-known substitutions such as $\tan(x/2)=t$. These substitutions were also found by birationally mapping the circle parametrized by $(\cos(x),\sin(x))$ onto the line $t\in\mathbb R$.

Answer 3

I think that it is necessary to adjust the formula solutions.

For the equation: $2X^2+2Y^2=Z^2+R^2$

Solutions can be written.

$X=p^2+2(t+k)ps-(t^2-2tk+k^2)s^2$

$Y=2(t+k)ps+2(t^2+2tk+k^2)s^2$

$Z=p^2+2(t+k)ps+(3t^2+6tk-k^2)s^2$

$R=p^2+2(t+k)ps+(3k^2+6tk-t^2)s^2$

For the system of equations:

$\left\{\begin{aligned}&X+Y+Z+R=0\\&2X^2+2Y^2=Z^2+R^2\end{aligned}\right.$

Solutions have the form:

$X=p^2+2ps-2s^2$

$Y=p^2-2ps-2s^2$

$Z=4s^2$

$R=-2p^2$

$p,s,t,k$ - integers of any sign.

Answer 4

Here is a low tech solution (it is the same answer with slightly different notation)

The solutions to $2(a^2+b^2)=c^2+d^2$ and $a+b=c+d$ are essentially $$c,d=4P^2,-2Q^2,a,b=2P^2\pm 2PQ-Q^2 $$ in that all solutions arise from these by possibly multiplying through by a constant $m \in \mathbb{Z}$. To have $\gcd(a,b,c,d)=1$ we should have $$\gcd(2P,Q)=1.$$ For a connection to Gaussian integers see the end.

The request was for $a+b=-(c+d)$ but that requires just changing the signs.

Here is (a sketch of) one way to arrive at that solution. Since we have $(a,b,c,d)=(1,1,2,0)$ lets look for other rational solutions with $a+b=c+d=2.$ We can then scale up to get (all) the integer solutions. So we seek $$2(a^2+b^2)=c^2+d^2 \text{ with }a,b=1\pm u\text{ and }c,d=1\pm v.$$ This simplifies to $$2(1+u^2)=1+v^2 \text{ i.e. }v^2=1+2u^2.$$

We can restrict this curve $v=\sqrt{1+2u^2}$ to $u,v \ge 0$ starting at the point $X=(0,1).$ If we take the line $v=1+ku$ of (rational) slope $k$ through $X$ it will intersect the curve at two points, $u=0$ and, after some algebra, $$u=\frac{2k}{2-k^2 }\text{ and }v=\frac{2+k^2}{2-k^2 }.$$ First substitue these values into $$a,b=1\pm u\text{ and }c,d=1\pm v.$$ Then set $k=\frac{P}{Q}$ to get expressions for $a,b,c,d$ all with denominator $P^2-2Q^2$. Clear the denominators to get the solution given above.

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For example with $k=\frac{2}{3}$ we get $u,v=6/7, 11/7$ leading to the rational solution $(a,b,c,d)=(1/7,13/7,-4/7,18/7)$ and thus the integer solution $(1,13,-4,18).$

ASIDE: Since $(a+bi)(a-bi)=a^2+b^2,$ the form of this problem suggests to me that there might be an approach via Gaussian integers (where there is a unique factorization up to units.) I did not fully work one out but the following is suggestive: $$a+bi=(1+i)(Q+P+Pi)(Q-P+Pi)$$ $$c+di =(1+i)^2(Q+P-Pi)(Q-P+Pi)$$ It might be enlightening to examine the variation $m(a^2+b^2)=c^2+d^2$ for $m=5,10,13$ or other values with all odd prime divisors of the form $4q+1$.