Parameterization of solutions for $a^2+b^2=c^2+d^2=2f^2$ with integer solutions

Question

I have little to no experience with Diophantine equations, and I have run across this problem while experimenting with magic squares.

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

a parameterization for $x^2+y^2=z^2+w^2$, can be plugged into a solution for $a^2+b^2=2c^2$ but this equation does not contain all integer values, and after experimenting, it misses most of them.

given $a^2+b^2=c^2+d^2=2f^2$, all integer solutions can be found for $a^2+b^2=2f^2$ by paramteterizing it into the following

$a = m^2+2mn-n^2$

$b = n^2+2mn-m^2$

$c = o^2+2op-p^2$

$d= p^2+2op-o^2$ this leads to the equation $o^2+p^2=m^2+n^2$, which all integer solutions can be found with the following parameterization

$m = ij + uv$

$n = jv - ui$

$o = ij - uv$

$p = jv + iu$

Given $m$, $n$, $i$, $j$, I can construct my original $x$, $y$, $z$, $w$ and a like this

$x = n^2+2mn-m^2,$

$y = m^2+2mn-n^2$

$z = o^2+2op-p^2$,

$w = p^2+2op-o^2$

$a = o^2+p^2$

And this gives me infinite solutions for any integers $i$, $j$, $u$ and $v$, but not every every solution to the equation, for example, it does not find $(127,97,58,74,113)$

I can create the construction of the $a^2+b^2=c^2+d^2$ using a difference of squares to factor it out into $op=uv$ and then create a parameterization for my variables, using a composite number $C$ created from even powers of primes, but that leaves me with $2f^2-a^2-c^2$ or $(f-a)(f+a)+(f-c)(f+c)$ and I'm unsure how to manipulate this.

Answer 1

HINT.- The general solution of $x^2+y^2=z^2+w^2$ it is known to be function of four parameters $$x=mt+ns\\y=ms-nt\\z=mt-ns\\w=ms+nt$$ On the other hand the general solution of $X^2+Y^2=2Z^2$ is function of two parameters and is given by $$X=\alpha^2-\beta^2+2\alpha\beta\\Y=\alpha^2-\beta^2-2\alpha\beta\\Z=\alpha^2+\beta^2$$ Because $(mt+ns)^2+ (ms-nt)^2=(m^2+n^2)(s^2+t^2)$ and we need $$(m^2+n^2)(s^2+t^2)=2(\alpha^2+\beta^2)$$if we choice $$m=\alpha_1^2-\beta_1^2+2\alpha_1\beta_1\\n=\alpha_1^2-\beta_1^2-2\alpha_1\beta_1$$and $s,t$ from a Pythagorean triple giving $s^2+t^2=\gamma^2$ we get this way $$(m^2+n^2)(s^2+t^2)=2(\alpha_1^2+\beta_1^2)\gamma^2=2((\alpha_1\gamma)^2+(\beta_1\gamma)^2)$$ so we have a parameterization of the diophantine $$a^2+b^2=c^2+d^2=2f^2$$ which obviously is not a general solution of the equation because it is function of the parameters $$\alpha_2=\alpha_1\gamma\\ \beta_2=\alpha_2\gamma\\\text { and }\gamma$$ NOTE.-I don't believe there is a general solution in part because it involves two distinct diophantine equations.

Answer 2

Earlier I had provided an answer having degree eight polynomial. This answer is a degree six polynomial with corresponding less number of terms & is shown below:

$a=(4 n^6 + 8 n^5 + 10 n^4 + 12 n^3 + 8 n^2 + 4 n + 1)$

$b=(4 n^6 + 16 n^5 + 22 n^4 + 20 n^3 + 12 n^2 + 4 n + 1)$

$c=(4 n^6 + 8 n^5 + 2 n^4 - 12 n^3 - 16 n^2 - 8 n - 1)$

$d=(4n^6+16n^5 + 30 n^4 +28 n^3 +12 n^2 - 1)$

$f=(2 n^2 + 2 n + 1) (2 n^4 + 4 n^3 + 4 n^2 + 2 n + 1)$

Hence,

$(a^2+b^2)=(c^2+d^2)=2(f)^2$

Where for n=1:

$(a,b,c,d,f)=(47,79,23,89,65)$

Answer 3

There is a parametric solution & is shown below:

a=$(7 k^8 + 2 k^7 + 41 k^6 + 44 k^5 + 92 k^4 + 136 k^3 + 92 k^2 + 112 k + 16)$

b=$(k^8 - 14 k^7 + 23 k^6 - 68 k^5 + 92 k^4 - 88 k^3 + 164 k^2 - 16 k + 112)$

c=$(k^8 + 14 k^7 + 23 k^6 + 68 k^5 + 92 k^4 + 88 k^3 + 164 k^2 + 16 k + 112)$

d=$(7 k^8 - 2 k^7 + 41 k^6 - 44 k^5 + 92 k^4 - 136 k^3 + 92 k^2 - 112 k + 16)$

f=$(k^4+5k^2+4)(5k^4+16k^2+20)$

And for, $(k=1)$ we get:

$(a,b)^2=(c,d)^2=2(f)^2$

$(271,103)^2=(289,23)^2=2(205)^2$