2
$\begingroup$

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

$$x^2+y^2+z^2=a^2$$

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?

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.