# 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? – Daniel W. Farlow Apr 17 '16 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 – KingJ Apr 17 '16 at 23:30
• Also took $m^3 + n^3 = (m+n)(m^2 -mn + n^2)$ – KingJ Apr 17 '16 at 23:31
• Thought of Lifting The Exponent, haven't worked deeply into this idea yet – KingJ Apr 17 '16 at 23:33
• You're about the prime factors: $2015=5\cdot13\cdot 31$. – Bernard Apr 18 '16 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 $7$.