Probably duplicate but I don't find: I'd like to solve the diophantine equation


which has solutions, by exemple $1^2+2^2+2^2=3^2$ or $2^2+3^2+6^2=7^2$. Every such solution gives a rational point on the unit sphere.

Is there a complete description of the solutions such as for pythagorician triplet?

  • 1
    $\begingroup$ See Wikipedia. $\endgroup$ – fuglede Jan 21 '14 at 14:50
  • $\begingroup$ I'd ask you to try @Piezas's collection! $\endgroup$ – Balarka Sen Jan 21 '14 at 20:30

We'll start with Pythagorean triples to see the pattern. The complete rational solution to $x_1^2+x_2^2 = y_1^2$ has the form,

$$((a^2-b^2)t)^2+(2abt)^2 = ((a^2+b^2)t)^2\tag{1}$$

where $t$ is a scaling factor.

Proof: For any solution where $x_1+y_1 \neq 0$ , one can always find rational {$a,b,t$} using the formulas $a,b,t = x_1+y_1,\; x_2,\; \frac{1}{2(x_1+y_1)}$.

Similarly, for $x_1^2+x_2^2+x_3^2 = y_1^2$, it is,

$$((a^2-b^2-c^2)t)^2+(2abt)^2+(2act)^2 = ((a^2+b^2+c^2)t)^2\tag{2}$$

Proof: One can always find rational {$a,b,c,t$} using $a,b,c,t = x_1+y_1,\; x_2,\; x_3,\; \frac{1}{2(x_1+y_1)}$.

and so on for $n$ squares. See also Sums of Three Squares for more.

  • $\begingroup$ Bautiful identity! It seems to me that the proof doesn't cover the integer case. Strange that for the classic case $x_1^2+x_2^2=y_1^2$ there's an "geometric" approach with $\mathbb{Z}[i]$ but there don't seem to be something similar for this case but we have still a good description of the solutions. $\endgroup$ – Macadam Jan 22 '14 at 14:54
  • $\begingroup$ I think there is geometric approach. Consider that $x^2+y^2=1$ describes a circle. Demjanenko studied $a^4+b^4+c^2=1$ as a pencil of conics, and Elkies specialized this to $a^4+b^4+c^4=1$ as a pencil of curves of genus one. $\endgroup$ – Tito Piezas III Jan 22 '14 at 20:41
  • $\begingroup$ I'm curious… for which $n$ does the complete rational solution [in rational parameters] have the same form as the complete integer solution [with integer parameters]? Clearly, $n=2$ does, and $n=3$ does not — see the famous Lebesgue Three-Square Identity (e.g. on your site). $\endgroup$ – Kieren MacMillan Sep 8 '14 at 21:30

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