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 $abcdefg\neq 0$ ?

Answer 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.

Answer 2

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

Answer 3

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