# Congruence - Number Theory

Prove that $2005^{2005}$ is not the sum of two perfect cubes.

I have looked at some mods but none have given me anything useful as of yet.

I looked at the usual mods such as $4, 5, 7, 11, 13$ but didn't see a way to use these with cubes.

I factored $m^3 + n^3$ but could not find anything useful for the problem.

I would like a proof using mods if possible as this is an easy olympiad problem and I don't want to over complicate it.

• Which ones did you look at and why were they not useful? Commented Apr 17, 2016 at 23:25
• Prime composition of 2015 is 5*401, looked at primes so the usual 4,5,11,13 etc. They were not useful because perfect cubes are harder to use is this way than even powers Commented Apr 17, 2016 at 23:30
• Also took $m^3 + n^3 = (m+n)(m^2 -mn + n^2)$ Commented Apr 17, 2016 at 23:31
• Thought of Lifting The Exponent, haven't worked deeply into this idea yet Commented Apr 17, 2016 at 23:33
• You're about the prime factors: $2015=5\cdot13\cdot 31$. Commented Apr 18, 2016 at 0:52

Note that $2005\equiv3$ in modulo $7$. Since $2005\equiv1$ in modulo $6$, you can work out that $2005^{2005}\equiv3$ in modulo $7$.
A perfect cube can only take values $1$, $0$ and $-1$ in modulo $7$. Therefore sum of two perfect cubes, namely sum of two of these numbers, can never be equivalent to $3$ in modulo $7$.