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

Question

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?

Answer 1

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!!

Answer 2

HINT.- You have $8\equiv-1\pmod9$. Besides the same property goes for $n\equiv5\pmod9$, I mean

If $n\equiv5\pmod9$ then $n$ cannot be written as the sum of three cubes.

The proof is equal to the same as you have made for $n\equiv4\pmod9$