Prove Integer Solution for $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3=k$

[anemone]

Prove that the equation $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3=k$ has an integer solution for any integer $k$.

 

[anemone]

Hint:

Spoiler

Note that $k=6n$ can be represented as a sum of four cubes.

 

[magneto1]

Spoiler

The hint follows from:
\[
(x+1)^3 + (x-1)^3 + (-x)^3 + (-x)^3 = 6x.
\]
Moroever, we have that under $\pmod 6$,
$(\pm 1)^3 \equiv \pm1$, $(\pm 2)^3 \equiv \pm 2$,
$3^3 \equiv 3$. We can choose any of the two of the cubes to form
$6x + k$ where $k = 1..5$.

 

[anemone]

Spoiler

The hint follows from:
\[
(x+1)^3 + (x-1)^3 + (-x)^3 + (-x)^3 = 6x.
\]
Moroever, we have that under $\pmod 6$,
$(\pm 1)^3 \equiv \pm1$, $(\pm 2)^3 \equiv \pm 2$,
$3^3 \equiv 3$. We can choose any of the two of the cubes to form
$6x + k$ where $k = 1..5$.

You're right magneto, well done and thanks for participating!:)