-1

Quite simply turned out to solve this Diophantine equation, when he made the assumption that the solutions of these equations symmetric.

So given this equation:

$$x^3+y^3+z^3+q^3+k^3=xyz+xyq+xyk+xzq+xzk+xqk+yzq+yzk+yqk+zqk$$

And symmetric solution is quite simple written.

$$x=25s^2+10ps+5p^2$$

$$y=10s^2+10ps+4p^2$$

$$z=20s^2+10ps+2p^2$$

$$q=5s^2+3p^2$$

$$k=15s^2+p^2$$

$s,p$ - integers of any sign.

The question is. This equation only symmetric solution? If not, what should be the idea for the solution of this equation?

individ
  • 4,301
  • This may be related if some one cares:http://math.stackexchange.com/questions/1588122/describe-the-nonzero-integer-solutions-to-the-equation-a3-b3-c3-d3 – NoChance Jan 07 '16 at 04:03
  • I'm sorry for a stupid comment but what is meant by 'symmetric solutions'? –  May 13 '17 at 14:39
  • @AknazarKazhymurat obtaining some private solutions using symmetric substitution! – individ May 13 '17 at 15:05

1 Answers1

0

In General, for any equation like this:

$$x^3+y^3+z^3+q^3+d^3+k^3=xyz+xyq+xyd+xyk+xzq+xzd+xzk+xqd+xqk+$$

$$+xdk+yzq+yzd+yzk+yqd+yqk+ydk+zqd+zqk+zdk+qdk$$

Symmetric solution can be written:

$$x=(3a^2+5b^2)p^2+5(9a^2+15b^2)s^2$$

$$y=80abps+200(a^2+5b^2)s^2$$

$$z=(a^2+15b^2)(p^2+15s^2)$$

$$q=5(a^2\pm2ab+5b^2)p^2+10(\pm5a^2+2ab\pm25b^2)ps+25(5a^2\mp6ab+25b^2)s^2$$

$$d=2(2a^2\pm5ab+5b^2)p^2+50(\pm{a^2}+2ab\pm5b^2)ps+10(31a^2\mp15ab+140b^2)s^2$$

$$k=2(a^2\pm5ab+10b^2)p^2+50(\pm{a^2}+2ab\pm5b^2)ps+10(28a^2\mp15ab+155b^2)s^2$$

$a,b,p,s$ - integers, any sign.

This is not clear. Quite limited symmetric solutions or need to look for others? And what way of calculation for this would be most effective?

individ
  • 4,301