why a³+b³+c³ is never 4 or 5 modulo 9 ? (all integers)

Question

could somebody help and inform why this is the case? Then I will have to understand of what form is a³+b³. a³ according to Fermat is of the form 1 modulo 3 (if a is not divisible by 3). Thanks.

Answer 1

Because $n^3\equiv0$ or $\pm1\pmod 9$.

Answer 2

Note that an integer n ≡ 0,1,2,3....,8 (mod 9)

So $n^3$ ≡ $0^3$,$1^3$,$2^3$....,$8^3$ mod 9

Or $n^3$ ≡ 0,1,8,27,64,343,512 (mod 9)

Or $n^3$ ≡ 0,1,8 (mod 9)

If u know some basics in congruences , u will see that 8 (mod 9) can be written as (-1) (mod 9)

So $n^3$ ≡ 0,1,-1 (mod 9)

Or $n^3$ ≡ 0, ±1 (mod 9)

So $a^3 + b^3 + c^3$ ≡ 0 + 0 + 0 (mod 9) , or 0 + 1 + 0 (mod 9) , or 1 + 0 + 0 (mod 9) , or 1 + 0 + (-1) (mod 9) and so on .

Check Every case and you will find $n^3$ is never ≡ 4 or 5 (mod 9)