Parametrization of integer solutions of the equation $a^2+b^2=c^2+d^2=2x^2$

Question

I need the general form of integer solutions to this equation $$a^2+b^2=c^2+d^2=2x^2$$ Here is my partial solution:-

The parametrization of the integer solutions of the equation $$p^2+q^2=2y^2$$ is the following:- $$p=k(m^2-n^2+2mn)$$ $$q=k(m^2-n^2-2mn)$$ $$y=k(m^2+n^2)$$ where $k,m,n\in\mathbb Z$. The proof can be seen here.

If we have two triplets of that satisfy the equations $p^2+q^2=2y^2$ and $r^2+s^2=2z^2$, we can multiply $p$, $q$ and $y$ with $z$ to get $$(pz)^2+(qz)^2=2(yz)^2$$ Similarly we can multiply $c$, $d$ and $z$ with $y$ to get $$(ry)^2+(sy)^2=2(yz)^2$$ Combining the two equations, we get $$(pz)^2+(qz)^2=(ry)^2+(sy)^2=2(yz)^2$$ If we set it like the following:- $$a=pz$$ $$b=qz$$ $$c=ry$$ $$d=sy$$ $$x=yz$$ we get integer solutions to $a^2+b^2=c^2+d^2=2x^2$ We can replace $p$ with $k(m^2-n^2+2mn)$, $q$ with $k(m^2-n^2-2mn)$ and $y$ with $k(m^2+n^2)$. Similarly, we can replace $r$ with $l(u^2-v^2+2uv)$, $s$ with $l(u^2-v^2-2uv)$ and $z$ with $l(u^2+v^2)$ and adding a scale factor $t$ to get $$a=k(m^2-n^2+2mn)l(u^2+v^2)t$$ $$b=k(m^2-n^2-2mn)l(u^2+v^2)t$$ $$c=k(u^2-v^2+2uv)l(m^2+n^2)t$$ $$d=k(u^2-v^2-2uv)l(m^2+n^2)t$$ $$x=k(u^2+v^2)l(m^2+n^2)t$$ This is a nice set of integer solutions to the equation $a^2+b^2=c^2+d^2=2x^2$. But I don't know if every integer solution will be in the form given above. If not, what is the parametrization of integer solutions to the equation that doesn't miss any integer solution? I also want the proof too, just to convince myself that the answer is true.

Edit:I have found from the comments that my set of equations misses a lot of solutions. So can somebody come up with a parametrization for this? (Where it is guaranteed that $a,b,c,d,x\in\mathbb Z$)

Update:I realized that my set of equations would miss some solutions where $k(m^2+n^2)$ and $l(u^2+v^2)$ are not relatively prime. So, instead of multiplying $p$, $q$ and $y$ with $z$ to get $(pz)^2+(qz)^2=2(yz)^2$ and multiplying $c$, $d$ and $z$ with $y$ to get $(ry)^2+(sy)^2=2(yz)^2$, we should multiply $p$, $q$ and $y$ with $\frac{LCM(y, z)}{y}$ and multiply $c$, $d$ and $z$ with $\frac{LCM(y, z)}{z}$ to get $$\left(\frac{pLCM(y, z)}{y}\right)^2+\left(\frac{qLCM(y, z)}{y}\right)^2=2(LCM(y, z))^2$$ and $$\left(\frac{rLCM(y, z)}{z}\right)^2+\left(\frac{sLCM(y, z)}{z}\right)^2=2(LCM(y, z))^2$$ Putting everything in terms of $m, n, u$ and $v$, we get

$$a=tk(m^2-n^2+2mn)\left(\frac{LCM(k(m^2+n^2),l(u^2+v^2))}{k(m^2+n^2)}\right)$$

$$b=tk(m^2-n^2-2mn)\left(\frac{LCM(k(m^2+n^2),l(u^2+v^2))}{k(m^2+n^2)}\right)$$

$$c=tl(u^2-v^2+2uv)\left(\frac{LCM(k(m^2+n^2),l(u^2+v^2))}{l(u^2+v^2)}\right)$$

$$d=tl(u^2-v^2-2uv)\left(\frac{LCM(k(m^2+n^2),l(u^2+v^2))}{l(u^2+v^2)}\right)$$

$$x=t(LCM(k(m^2+n^2),l(u^2+v^2)))$$

Now, these set of equations include solutions like $(|a|,|b|,|c|,|d|,|x|)=(5,35,17,31,25)$ and $(|a|,|b|,|c|,|d|,|x|)=(1,7,5,5,5)$ which were missing with my earlier set of equations. But, I am still not sure if this set of equations misses any solutions. So, if this set of equation misses any solutions, I need the true parametrization of the integer solutions of the equation that doesn't miss any solution and its proof. Or if this is an actual parametrization of the equation that includes every integer solution, I need the proof for that.

Answer 1

For such a system of equations and, for example, other similar ones.

https://artofproblemsolving.com/community/c3046h2985976_a_system_of_similar_nonlinear_equations

There is a standard solution method. We write down the solution of the equations as follows.

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

$$a=p^2+2ps-s^2$$ $$b=-p^2+2ps+s^2$$ $$c=k^2+2kt-t^2$$ $$d=-k^2+2kt+t^2$$ The parametrization has already been written down and only this equation needs to be solved. $$x=p^2+s^2=k^2+t^2$$

The formula for solving this equation can be found there.

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

Answer 2

Partial answer. Need to finish.

Gaussian Integers might help. A Gaussian Integer is a a complex number with integer real and imaginary parts.

A Gaussian Integer $z$ is a prime Gaussian Integer iff It is a standard prime $p$, where $p\equiv 3 \pmod{4}$, that same prime times $i$ or the squared modulus of $z$ is a prime. Consequently 5 is not prime, for $5=(2+i)(2-i)$. 2 is an ambiguous case since $i(1-i)^2=(1+i)(1-i)$. The prime factorization of a Gaussian Prime is $z=u(i+1)^{e_0}p_1^{e_1}p_2^{e_2}\dots $ where $u \in \{1,-1,i,-i\}$

If $a^2+b^2=c^2+d^2=2x^2$ and $a\ne b$, $c\ne d$ .

$(a+bi)(a-bi)=(c+di)(c-di)=(1+i)(1-i)x^2$

$2x^2$ is not prime so neiter are $a+bi$ and $c+di$. Let $a+bi=pq$ for Gaussian Integers $p$ and $q$.

Then $pq\bar{p}\bar{q}=(1+i)(1-i)x^2$

We also have that $rs=c+di\ne pq$

$a+bi=pq$, $c+di=p\bar{q}$

$pq=(t+yi)(w+zi)=(tw-yz)+i(tz+wy)$

$p\bar{q}=(t+yi)(w-zi)=(tw+yz)+i(wy-zt)$

$a=wt-yz. b=tz+wy.$

$c=wt+yz. d=wy-zt$

$(a+c)/2=wt$. $(b+d)/2=wy$

$y=t\frac{(b+d)}{(a+c)}$

$w=(a+c)/(2t)$

$z=(b-d)/2t$

$(1\pm i)|(wt-yz)+i(tz-wy)=a+bi$

This might suggest a parameterization.