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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.

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2 Answers2

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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.

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

individ
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    I don't quite understand the last part. Also, the second link give multiple set of solutions. Does that mean you can choose whichever set of solutions you like? –  Mar 14 '23 at 10:24
  • @Hisham I think the parametrization earlier on that question is just a tiny bit easier to use: $(a,b,c,d)=(pr+qs, qr-ps, pr-qs, ps+qr)$ – Eric Snyder Mar 15 '23 at 01:10
  • Actually, this doesn't work, as it still won't catch situations where one of the sequences would be parametrized by non-integers. – Eric Snyder Mar 16 '23 at 06:08
  • @EricSnyder It works. This is the only method for solving such systems of equations. The meaning there is simple. Reduce the solution of a system of equations to the solution of a single equation. A difficult task turns into an extremely simple one. In addition, other systems with a different form can be solved. – individ Mar 16 '23 at 06:13
  • @individ Please show how this method generates the solution $(5,35,17,31,25)$, which requires ${(k,t),(p,s)} = {(\sqrt5,2\sqrt5),(3,4)}$, from integers. Or, I suppose, show that $5$ and $35$ can be found as $2mn+m^2-n^2, 2mn-m^2+n^2$ with integer $m,n$. This will be difficult because $2xy+x^2-y^2$ in the integers in general is restricted to $\pm 1 \pmod 8$. – Eric Snyder Mar 16 '23 at 07:28
  • @EricSnyder I am constantly required to do some arithmetic. Forget about arithmetic - look at the formulas. Any parametrization is written taking into account multiple solutions. That is, mutually simple solutions are not all solutions. It is necessary to divide the solutions by a common divisor. To simplify the recording, this is usually excluded and such a record is not written. Very often they forget about it and there are questions that you ask me. – individ Mar 16 '23 at 08:28
  • @individ So what you're saying is, your parametrization doesn't cover all solutions, much as was the case with OP's original attempt? The reason this method misses those solutions is that we would need a parametrization for $i(m^2+n^2)=j(p^2+q^2)=x$, which doesn't seem to be readily found. It has nothing to do with the parametrization producing examples that require division by the GCD. – Eric Snyder Mar 16 '23 at 08:37
  • @EricSnyder Any parametrization is written with a common multiplier. And taking into account the total multiplier, just get all the solutions. This recording option looks more compact. Because if I write the solution of the equation that you wrote, then the parametrization formula will be extremely cumbersome. And besides, it will be impossible to write the formula using the approach that I described. It will be necessary to solve the equation of 4 degrees. – individ Mar 16 '23 at 09:09
  • @individ You are creating a really biased contradiction here with "Forget about arithmetic - look at the formulas." - for the formulas to be confirmed to give an identity, one has to do arithmetic anyway. But arithmetic or not, the question is really if one may use irrational number for any parameter to get a solution in integers that can not be produced by rationals - it is neither proved of disproved. As a matter of fact, you are not saying anywhere to what set the parameters belong. – Dávid Laczkó Mar 17 '23 at 08:57
  • @DávidLaczkó I have been asked this question many times in various combinations. It happens that you need to write down the solution of the equation in a general form and it may look like you need to reduce by a common factor. http://math.hashcode.ru/questions/249507/теория-чисел-доказать-что-a-b-c-d-составное-число?страница=1&focusedAnswerId=249550#249550

    Or, for example, a Pythagorean triple can contain 3 parameters. But here the claim is that they require the form of the law of nature in the form in which they like.

     – individ Mar 17 '23 at 10:16