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We now that $a^3 +b^3=c^3$ has no solution if $a,b,c\in\mathbb{N}$(thus non of $a$, $b$ or $c$ can be zero). Well I want to know whether this can be proven using congruency (Like how we can prove that there is no sum of square of the can be of the form $4n+3$).

This is how I started:

for any $a$, $a^3\equiv-1,0,1\pmod{7}$. And the same implies for $b$. Then I took all the $6$ cases and tried to eliminate unwanted cases. I could not go ahead. Is it possible to prove it this way?

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    $\begingroup$ As $0^3+b^3=b^3$ you certaily cannot eliminate the case that $7\mid a$. $\endgroup$ – Hagen von Eitzen Jun 15 '16 at 8:12
  • $\begingroup$ @HagenvonEitzen, I have said that they belong to $\mathbb{N}$. As per the normal definition $0\notin \mathbb{N}$ $\endgroup$ – Coherent Sheaf Jun 15 '16 at 8:17
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    $\begingroup$ @AbhijitAJ Well, it could be that $a =0 \pmod{7}$: you wrote it by yourself. $\endgroup$ – Crostul Jun 15 '16 at 8:28
  • $\begingroup$ I won't say it's impossible to prove it via congruences, but I'm sure it has never been done, and I'm sure lots of people have tried. $\endgroup$ – Gerry Myerson Jun 15 '16 at 9:00
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    $\begingroup$ The classic Ramsey-theoretic theorem of Issai Schur was actually motivated by trying to understand this number-theoretic question. One of his results is that for any prime $p > 17$, there always exists a modular solution to $a^3+b^3\equiv c^3 \pmod p$ with $a,b,c$ all non-zero. This leaves only a small number of cases to check manually (and of course only primes congruent to $1$ mod $3$ are of any interest). $\endgroup$ – Erick Wong Jun 22 '16 at 6:02
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A simple argument by congruency cannot establish that $a^3+b^3=c^3$ has no non-trivial solutions. In particular, notice the following identity: $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ The significance of this is that, suppose we choose $a$ and $b$ such that $a+b\equiv 0 \pmod n$. This implies that $a^3+b^3\equiv 0\pmod n$, which is a problem, since $0$ is a cube mod any $n$. For a proof, we would need to show that $a^3+b^3\equiv c^3\pmod n$ has only solutions where $n$ divides all of $a,\,b,\,c$, but our method gives plenty of examples where this is not true. One might more succinctly note the particular case $$(n-1)^3+(n+1)^3\equiv n^3 \pmod n$$ for any $n$.

That said, it might be possible to find ways to restrict the possibilities of $a$ and $b$ usefully mod $n$. A useful fact to this is that if $p$ is a prime of the form $6m-1$, then every number is a cube mod $p^k$ and if $p$ is a prime of the form $6m+1$, then a third of the numbers are cubes mod $p^k$. These facts follow from the fact that the multiplicative group mod a prime power is the cyclic group of order $\varphi(p^k)$. We get that looking at the equation mod powers of primes of the form $6m+1$ is the only possible way to be productive.

In the particular case of $n=7$, it's worth noting that the observation that the only cubes are $-1,\,0,\,1$ allows us to consider only solutions of the form $$(7a'+1)^3+(7b')^3+(7c'-1)^3=0$$ It would not surprise me if the non-existence of solutions to $a^3+b^3=c^3$ can be proven by following this path, though I am not aware of any such proof.

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