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Is a complete parametrization of primitive solutions to the equation $$a^2+b^2+c^2=2d^2\qquad a,b,c,d \in \mathbb{Z}$$ known? A reference would be great.

Solutions to $a^2+b^2=c^2$ give solutions to the equation above, but I know that there are other solutions.

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Alright, this is from Jones and Pall (1939), I have a pdf if you wish to investigate.

Find all ways to write $$ m = t^2 + u^2 + 2 v^2 + 2 w^2, $$ with $m$ odd. There is no reason to consider even $m$ for this problem. Using material from pages 174-177, all primitive solutions of $$ 2 m^2 = x^2 + y^2 + z^2 $$ can then be written, up to order and signs, as $$ x = 4 tw+ 4 uv, $$ $$ y = t^2 - 2tu -u^2 +2v^2 +4vw - 2 w^2, $$ $$ z = t^2 + 2tu -u^2 +2v^2 -4vw - 2 w^2. $$ Since $m$ is odd, $2m^2 \equiv 2 \pmod 8,$ two out of three of $x,y,z$ must be odd, the other divisible by $4.$

Those below are primitive, that is $\gcd(x,y,z) = 1.$ In order to get all possible solutions, take a quadruple $(m;x,y,z)$ and multiply all four by any constant you like.

  m      x      y      z  
  1      0      1      1        t : 1  u : 0  v : 0  w : 0
  3      4      1      1        t : 1  u : 0  v : 0  w : 1
  5      0      7      1        t : 2  u : 1  v : 0  w : 0
  5      4      5      3        t : 1  u : 0  v : 1  w : 1
  7      4      9      1        t : 2  u : 1  v : 1  w : 0
  7      8      5      3        t : 2  u : 1  v : 0  w : 1
  9      4     11      5        t : 2  u : 1  v : 1  w : -1
  9      8      7      7        t : 1  u : 0  v : 0  w : 2
 11     12      7      7        t : 3  u : 0  v : 0  w : 1
 11      4     15      1        t : 1  u : 0  v : 2  w : 1
 11      8     13      3        t : 1  u : 0  v : 1  w : 2
 13      0     17      7        t : 3  u : 2  v : 0  w : 0
 13     12     13      5        t : 3  u : 0  v : 1  w : 1
 13     16      9      1        t : 2  u : 1  v : 0  w : 2
 13      8     15      7        t : 2  u : 1  v : 2  w : 0
 15     16     13      5        t : 2  u : 1  v : 2  w : 1
 15     20      7      1        t : 2  u : 1  v : 1  w : 2
 15      8     19      5        t : 3  u : 2  v : 1  w : 0
 17      0     23      7        t : 4  u : 1  v : 0  w : 0
 17     20     13      3        t : 3  u : 2  v : 1  w : 1
 17     24      1      1        t : 3  u : 0  v : 0  w : 2
 17      4     21     11        t : 3  u : 2  v : 1  w : -1
 17      8     17     15        t : 1  u : 0  v : 2  w : 2
 19     12     17     17        t : 1  u : 0  v : 0  w : 3
 19     12     23      7        t : 3  u : 0  v : 2  w : 1
 19     16     21      5        t : 4  u : 1  v : 0  w : 1
 19     24     11      5        t : 3  u : 0  v : 1  w : 2
 19      4     25      9        t : 4  u : 1  v : 1  w : 0
 21     16     25      1        t : 3  u : 2  v : 2  w : 0
 21     20     19     11        t : 4  u : 1  v : 1  w : 1
 21      4     29      5        t : 1  u : 0  v : 3  w : 1
 21      8     23     17        t : 2  u : 1  v : 2  w : -2
 23     12     25     17        t : 2  u : 1  v : 3  w : 0
 23     16     21     19        t : 3  u : 2  v : 1  w : -2
 23     24     19     11        t : 2  u : 1  v : 0  w : 3
 23     28     15      7        t : 3  u : 2  v : 2  w : 1
 23     32      5      3        t : 3  u : 2  v : 1  w : 2
 23      4     31      9        t : 3  u : 2  v : 2  w : -1
 25      0     31     17        t : 4  u : 3  v : 0  w : 0
 25     20     27     11        t : 2  u : 1  v : 3  w : 1
 25     20     29      3        t : 2  u : 1  v : 1  w : -3
 25     24     25      7        t : 3  u : 0  v : 2  w : 2
 25     28     21      5        t : 2  u : 1  v : 1  w : 3
 25     32     15      1        t : 4  u : 1  v : 0  w : 2
 25      4     35      3        t : 2  u : 1  v : 3  w : -1
 25      8     31     15        t : 4  u : 1  v : 2  w : 0
  m      x      y      z  

Note how very similar this is to

https://en.wikipedia.org/wiki/Pythagorean_quadruple#Parametrization_of_primitive_quadruples

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Look for a single rational solution to $x^2+y^2+z^2=2$. Say, $(x_0,y_0,z_0)=(-1,0,1)$.

Then take any three integers $(u,v,w)$ and seek solutions of form $(-1+tu,tv,1+tw)$. You get:

$$1-2tu+t^2u^2 + t^2v^2 + 1+2tw + t^2w^2=2$$

solving, excluding $t=0$, you get:

$$t=\frac{2(u-w)}{u^2+v^2+w^2}$$

Substituting that bach in, we get:

$$\begin{align} x&=-1+tu = \frac{-(u^2+v^2+w^2)+2(u-w)u}{u^2+v^2+w^2}&=&\frac{u^2-2uw-v^2-w^2}{u^2+v^2+w^2}\\ y&=tv &=& \frac{2(u-w)v}{u^2+v^2+w^2}\\ z&=1+tw &=&\frac{u^2+v^2-w^2 +2uw}{u^2+v^2+w^2} \end{align}$$

Then $$(a,b,c,d)=(u^2-2uw-v^2-w^2,2(u-w)v,u^2+v^2-w^2+2uw,u^2+v^2+w^2)\tag{1}$$

When $w=0$, this gives: $(a,b,c,d)=(u^2-v^2,2uv,u^2+v^2,u^2+v^2)$, which is the result you note that if $(a,b,c)$ is a Pythagorean triple then $(a,b,c,c)$ is a solution of your equation. (I had to pick the $(x_0,y_0,z_0)$ carefully to make that work.

The formula (1) should yield all integer solutions, up to constant multiples.

To ensure a primitive solution, you need that $u+v+w$ is odd and that $u-w$ and $v^2+2u^2$ are relatively prime. That is equivalent to $\gcd(2(u-w),u^2+v^2+w^2)=1$, which is saying that $t$ above is in lowest common denominator form.

As Will noted in comments, this doesn't give all primitive solutions directly, since:

$$9^2+4^2+1^2=2\cdot 7^2$$

is a solution, and there is no way to write $7$ as the sum of three squares.

Instead, we get: $21=4^2+2^2+1^2$, and we get a solution:

$$(27,12,3,21)$$ from my formula, which is obviously not primitive.

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  • $\begingroup$ In $x=$ and then in $a=$ the sign of $2uw$ should be negative. Thank you! $\endgroup$ – vuur Oct 28 '15 at 13:48
  • $\begingroup$ Fixed. You're welcome. $\endgroup$ – Thomas Andrews Oct 28 '15 at 13:51
  • $\begingroup$ Given any rational homogeneous quadratic polynomial $p(x_1,\dots,x_n)$ if you can find one solution to $p(x_1,\dots,x_n)=C$ then you can use this approach to find all rational solutions. $\endgroup$ – Thomas Andrews Oct 28 '15 at 14:46
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    $\begingroup$ right, in your (1), we cannot have $u^2 + v^2 + w^2 \equiv 7 \pmod 8,$ although we can have $d \equiv 7 \pmod 8.$ (9,4,1,7) $\endgroup$ – Will Jagy Oct 28 '15 at 18:16
  • $\begingroup$ found parametrization similar to en.wikipedia.org/wiki/… I will check with a computer program later or tomorrow. $\endgroup$ – Will Jagy Oct 29 '15 at 1:24

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