This is a repeat of a question asked here.
Show that if $n \equiv 4 (\mod 9)$, then $n$ cannot be written as the
sum of three cubes.
Solution: Any integer has least residue as either $0,1,2,3,4,5,6,7,8 (\mod 9)$
Now, $$0^3\equiv0 \pmod 9\\
1^3\equiv1 \pmod 9\\
2^3\equiv8 \pmod 9\\
3^3\equiv0 \pmod 9\\
4^3\equiv 1\pmod 9\\
5^3\equiv 8\pmod 9\\
6^3\equiv 0\pmod 9\\
7^3\equiv 1\pmod 9\\
8^3\equiv 8\pmod 9$$.
So,any integer cube is congruent to either $0,1 \text{or} 8 \pmod 9$
It is not possible to produce $4$ with the combination of $0,1,8$.
Closest we can get is $1+1+1=3$ or $0+0+8=8$.
My question (1) is this method correct?
Question (2) is: Why the answer to the linked question says that $m^3 \equiv 0, \pm1 \pmod9$? How the user got "$-1$" and why "$8$" is missing?
Best Answer
Yes your method is correct according to the question .
Regarding the second part of your question, You fetched $n'\equiv0(mod9)$
$n'\equiv1(mod9)$
$n'\equiv8(mod9)$
$\forall n'\in\mathbb Z $and n' is the sum of 3 cubes
Now you see
The actual definition of congruences is
If $ m\equiv n(mod a)$ $Then ,$ $m=a.q_1+r$
And $b=a.q_2+r$
Where $ a,q_1,q_2,r\in \mathbb Z$ Or in other words two numbers are said to be congruent modulo a iff they leave same remaiders when divided by a
Now you see we say If $m\equiv n(mod a)$,
Then $m\equiv n\pm k.a(mod a)$
PROOF:
Given:$m\equiv n(mod a)$
To show:$m\equiv n\pm k.a(mod a)$
proof:
If $m\equiv n(mod a)$. Then m and n leaves same remainder with a. Now,
$n=a.q_2+r$.
$\Rightarrow n\pm k.a=a.q_2+r\pm k.a$
$\Rightarrow n\pm k.a=a(q_2\pm k.a)+r$ where $k\in \mathbb Z$
Now , Since $q_2,k\in \mathbb Z$
And $q_2 \pm k.a\in \mathbb Z$ So we see now $n \pm a $ leaves same remainder as m and n So we can say,
$m\equiv n\pm k.a (mod a)$
Using this theorem as we prove what you have asked
You found
$n'\equiv 8(mod9)$
Therefore $n'\equiv 8-9(mod 9)$
So we get $n'\equiv -1(mod 9)$
So writing 8(mod 9) and -1(mod 9) is all the same.
So you are done!!