MHB Prove x^31+x^32+x^33+x^34+x^35=n has an integer solution for any integer n.

[lfdahl]

Prove, that the equation

\[x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 = n\]

has an integer solution for any integer $n$.

 

[kaliprasad]

Prove, that the equation

\[x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 = n\]

has an integer solution for any integer $n$.

Spoiler

We have $(t + 1)^3 + (t- 1)^3 = 2t^3 + 6t$
or $6t = (t+1)^3 + (t-1)^3 - t^3 - t^3 = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$
let us define $f(t) = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$ which is sum of of 4 cubes

the numbers are of the form
$6t = f(t)+ 0 $ sum of 5 cubes
$6t+1 = f(t) + 1^3$ sum of 5 cubes
$6t+2 = 6(t-1) + 8 = f(t-1) + 2^3$ sum of 5 cubes
$6t+3 = 6(t-4) + 27 = f(t-4)+ 3^3 $ sum of 5 cubes
$6t+4 = 6(t-10) + 64 = f(t-10) + 4^3$ sum of 5 cubes
$6t+5 = 6(t-20) + 125 = f(t-20) + 5^3$ sum of 5 cubes
as we have taken care of all cases we are done

 

[lfdahl]

Spoiler

We have $(t + 1)^3 + (t- 1)^3 = 2t^3 + 6t$
or $6t = (t+1)^3 + (t-1)^3 - t^3 - t^3 = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$
let us define $f(t) = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$ which is sum of of 4 cubes

the numbers are of the form
$6t = f(t)+ 0 $ sum of 5 cubes
$6t+1 = f(t) + 1^3$ sum of 5 cubes
$6t+2 = 6(t-1) + 8 = f(t-1) + 2^3$ sum of 5 cubes
$6t+3 = 6(t-4) + 27 = f(t-4)+ 3^3 $ sum of 5 cubes
$6t+4 = 6(t-10) + 64 = f(t-10) + 4^3$ sum of 5 cubes
$6t+5 = 6(t-20) + 125 = f(t-20) + 5^3$ sum of 5 cubes
as we have taken care of all cases we are done

You really cracked the nut with a splendid solution, kaliprasad! Thanks a lot for your participation!

 

[kaliprasad]

Spoiler

We have $(t + 1)^3 + (t- 1)^3 = 2t^3 + 6t$
or $6t = (t+1)^3 + (t-1)^3 - t^3 - t^3 = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$
let us define $f(t) = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$ which is sum of of 4 cubes

the numbers are of the form
$6t = f(t)+ 0 $ sum of 5 cubes
$6t+1 = f(t) + 1^3$ sum of 5 cubes
$6t+2 = 6(t-1) + 8 = f(t-1) + 2^3$ sum of 5 cubes
$6t+3 = 6(t-4) + 27 = f(t-4)+ 3^3 $ sum of 5 cubes
$6t+4 = 6(t-10) + 64 = f(t-10) + 4^3$ sum of 5 cubes
$6t+5 = 6(t-20) + 125 = f(t-20) + 5^3$ sum of 5 cubes
as we have taken care of all cases we are done

better solution that is closed form ( first 3 lines copied from above)

Spoiler

We have $(t + 1)^3 + (t- 1)^3 = 2t^3 + 6t$
or $6t = (t+1)^3 + (t-1)^3 - t^3 - t^3 = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$
let us define $f(t) = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$ which is sum of of 4 cubes

now $n^3-n= n(n^2-1) = n(n+1)(n-1)= (n-1)n(n+1) = 6\frac{(n-1)n(n+1}{6} $ Now $\frac{(n-1)n(n+1}{6}$ is integer as $(n-1)n(n+1)$ being multiple of 3 consecutive numbers is divisible by 3! or 6

so $6t+n = n^3 + 6t -(n^3-n) = n^3 + 6(t-\frac{(n-1)n(n+1)}{6}) = n^3 + f(t-\frac{(n-1)n(n+1)}{6})$ sum of one cube and 4 cubes or sum of 5 cubes

 

[lfdahl]

better solution that is closed form ( first 3 lines copied from above)

Spoiler

We have $(t + 1)^3 + (t- 1)^3 = 2t^3 + 6t$
or $6t = (t+1)^3 + (t-1)^3 - t^3 - t^3 = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$
let us define $f(t) = (t+1)^3 + (t-1)^3 + (-t)^3 + (-t)^3$ which is sum of of 4 cubes

now $n^3-n= n(n^2-1) = n(n+1)(n-1)= (n-1)n(n+1) = 6\frac{(n-1)n(n+1}{6} $ Now $\frac{(n-1)n(n+1}{6}$ is integer as $(n-1)n(n+1)$ being multiple of 3 consecutive numbers is divisible by 3! or 6

so $6t+n = n^3 + 6t -(n^3-n) = n^3 + 6(t-\frac{(n-1)n(n+1)}{6}) = n^3 + f(t-\frac{(n-1)n(n+1)}{6})$ sum of one cube and 4 cubes or sum of 5 cubes

Thanks again, kaliprasad!Yes, the latter solution is more compact in its closed form. Nice observation!