Describe the nonzero integer solutions to the equation $a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 =0$

Question

Can someone describe all the integer solutions to the above equation such that $abcdefg\neq 0$ ?

Here's one solution that I thought of $(-1,-1,-1,-2,-2,-2,3)$. Multiplying each term by $-1$ also works. — cheesyfluff, Dec 24 '15 at 20:58
@ NoChance, may you kindly outline the ways, at least that would be some progress. — Q_p, Dec 24 '15 at 21:12
Take a look at:https://sites.google.com/site/tpiezas/010, some of the expressed formulas could help. — NoChance, Dec 24 '15 at 22:00
In addition to my other note, the fact that the sum of cubes equals the square of the sum is interesting and may help you: http://arxiv.org/pdf/1306.5257.pdf — NoChance, Dec 24 '15 at 22:29

score 1 · Answer 1 · answered Dec 24 '15 at 21:56

1

I do not think there is a known way of describing all of them, but here you'll find some further reading: http://mathworld.wolfram.com/CubicNumber.html

answered Dec 24 '15 at 21:56

flawr

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

For the equation.

$$x_1^3+x_2^3+x_3^3+x_4^3+x_5^3+x_6^3=x_7^3$$

You can write a fairly simple formula.

$$x_1=t^2-3(k+s)(p+t)-3p^2+2u$$

$$x_3=t^2-3(p+s)(k+t)-3k^2+2u$$

$$x_5=t^2-3(p+k)(s+t)-3s^2+2u$$

$$x_2=2t^2+3((k+s)(p-t)-2pt)+3p^2+u$$

$$x_4=2t^2+3((p+s)(k-t)-2kt)+3k^2+u$$

$$x_6=2t^2+3((p+k)(s-t)-2st)+3s^2+u$$

$$x_7=3(t^2-2(p+k+s)t+u)$$

where,

$$u=3(p^2+k^2+s^2)$$

Cube certainly look nice, but I prefer to solve such equations. Look cumbersome, but the solution much simpler. The sum of the cubes and the amount of combinations.

edited Apr 13 '17 at 12:20

Community

answered Dec 25 '15 at 09:39

individ

You can always simplify expressions by using substitutions. – Tito Piezas III Jan 07 '16 at 03:51
Care to explain where did the substitution formula came from? – NoChance Jan 07 '16 at 04:05
1

@NoChance: It's just by inspection. If you look at his version 1, you can see how the expressions for the $x_i$ can easily be shortened by using the variable $u$. – Tito Piezas III Jan 07 '16 at 13:44
Thanks for the explanation. – NoChance Jan 07 '16 at 14:05

score -1 · Answer 3 · answered Jun 29 '17 at 05:43

Regarding equation, (a^3 + b^3 + c^3 + d^3 + e^3 + f^3 = g^3 )

The above equation has parametrization given below:

(a,b,c,d,e,f)^3=

[(6k^2+12k-18),(k^2-2k+49),(5k^2+38k+5),(8k^2+16k-24), (7k^2-14k-41),(10k^2+20k-30)]^3 = [(13k^2+22k+13)]^3= (g)^3

For k=2 we have:

(30,49,101,40,-41,50)^3=(109)^3

3 Answers3