how do you solve $a^2+b^2+c^2=d^3$

Question

let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$

Answer 1

It is a theorem that one can identically solve,

$$x_1^2+x_2^2+\dots+x_n^2 = (y_1^2+y_2^2+\dots+y_n^2)^k$$

for any positive integer n and k. Thus the kth power of n squares is itself the sum of n squares. For example, for $n =3$, we have,

$k=2:$

$$(a^2-b^2-c^2)^2+(2ab)^2+(2ac)^2 = (a^2 + b^2 + c^2)^2$$

$k=3:$

$$a^2(a^2 - 3b^2 - 3c^2)^2 + b^2(-3a^2 + b^2 + c^2)^2 + c^2(-3a^2 + b^2 + c^2)^2=(a^2 + b^2 + c^2)^3$$

and so on. See Theorem 1 at https://sites.google.com/site/tpiezas/004.

Answer 2

The simple formula for the equation:

$$x^2+y^2+z^2=q^3$$

You can write this:

$$x=3(p-k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$y=3(p-k+2t)(p^2+2k^2-2kt+2t^2)s^3$$

$$z=3(p+2k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$q=3(p^2+2k^2-2kt+2t^2)s^2$$

$p,k,t$ - integers asked us.

If there is a one simple solution, it will be necessary to reduce to the corresponding $s$.

There are other formulas. But they are bulky. Don't know whether it makes sense to write them.

Answer 3

For the equation:

$$x^2+y^2+z^2=r^3$$

Will make a replacement that formula was compact.

$$c=2(q-p-s)t$$

$$d=s^2+t^2-q^2-p^2+2p(q-s)$$

$$k=p^2+t^2-q^2-s^2+2s(q-p)$$

$$n=p^2+t^2+s^2-q^2$$

$$j=p^2+s^2+t^2+q^2-2q(p+s)$$

$p,s,t,q$ - integers asked us. Then decisions can be recorded.

$$x=dn^2+2cnj-dj^2$$

$$y=cj^2+2dnj-cn^2$$

$$z=k(n^2+j^2)$$

$$r=n^2+j^2$$

Answer 4

Refer to Gohierre de Longchamps's formula \begin{gather*} \left\{ \begin{split} a&=u\Big(\lambda\,\!u^2+\mu\,\!v^2-3\nu\,\!w^2\Big)\\ b&=v\Big(\lambda\,\!u^2+\mu\,\!v^2-3\nu\,\!w^2\Big)\\ c&=w\Big(3\lambda\,\!u^2+3\mu\,\!v^2-\nu\,\!w^2\Big)\\ d&=\lambda\,\!u^2+\mu\,\!v^2+\nu\,\!w^2 \end{split} \right.\\ \\ \Downarrow\\ \\ \Large{\lambda\,\!a^2+\mu\,\!b^2+\nu\,\!c^2=d^3} \end{gather*}

L'Intermédiaire des mathematiciens. 1902-1903. P311-312