If $n\equiv 4 \pmod 9$ then $n$ cannot be written as sum of three cubes?

Question

Show that if $n\equiv 4 \pmod 9$ then $n$ cannot be written as sum of three cubes.

This might be a silly question but I really don't see it? The thing I ended up was: let $n=a^3 + b^3 + c^3$, then we'll end up with $[a]^3+[b]^3+[c]^3=[4]$ in $Z_9$. I found several webpages and apparently this is quite "obvious".

Using the command $\tt{\backslash}\tt{pmod}$ gives good results for parenthesis modulo notation. — Travis Willse, Apr 22 '15 at 03:31

score 5 · Accepted Answer · answered Apr 22 '15 at 03:28

5

HINT Note that $m^3 \equiv 0, \pm1 \pmod9$.

answered Apr 22 '15 at 03:28

Adhvaitha

20,259

1

Aha, I see it now. This result is not "obvious" to me though.. :( – 3x89g2 Apr 22 '15 at 03:36
By the way how would you prove this statement? Just simply go through $0^3$ to $9^3$? – 3x89g2 Apr 22 '15 at 04:25
A more rigorous explanation and proof is here : https://math.stackexchange.com/a/1523147/1311 – Abel Mar 01 '20 at 03:20

If $n\equiv 4 \pmod 9$ then $n$ cannot be written as sum of three cubes?

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