Positive solutions to this exponential diophantine Find positive integral solutions to :
$2013^p$ $+$ $2014^q$ = $2015^r$
Attempt:
$2013^p$ $+$ $(2013 + 1)^q$ $=$ $($2013$ + $2$)^r$
Expanding,
$2013^p$ + $2013^q$ + $1$ + $2013Q$ = $2013^r$ + $2^r$ + $2013R$
Here R and Q are some constants, via binomial theorem expansion of $(a+b)^n$ involves aⁿ, bⁿ and something divisible by ab
Anyways, looking at modulo 2013
$0$ + $0$ + $1$ + $0$ = $0$ + $2^r$ + $0$
Or, 1 congruent to $2^r$ modulo 2013
Now how do I solve this?
 A: It's usually not helpful to take the equation modulo anything other than a prime or a prime power. We can factor $2013$ as $3 \cdot 11 \cdot 61$. Taking the equation modulo $61$ tells us that $0^p + 1^q \equiv 2^r \pmod{61}$, or $2^r \equiv 1 \pmod{61}$, which only happens if $r$ is a multiple of $60$. When $r$ is a multiple of $60$, we have $2^r \equiv 1 \pmod 3$ and $2^r \equiv 1 \pmod{11}$ as well, so we gain no other information from considering those moduli.
We can learn a bit from prime factors of $2014$ and $2015$:


*

*Taking the equation modulo $19$ (a small prime factor of $2014$), we get $(-1)^p + 0^q \equiv 1^r \pmod{19}$, from which we learn that $p$ is even.

*Taking the equation modulo $31$ (a small prime factor of $2015$), we learn that $(-2)^p + (-1)^q \equiv 0^r \pmod{31}$, or $4^{p/2} + (-1)^q \equiv 0 \pmod{31}$. No power of $4$ is congruent to $-1$ modulo $31$, and we can only satisfy this if $p/2$ is a multiple of $5$ and $q$ is odd. 

*Taking the equation modulo $13$ (another small prime factor of $2015$),  we get $(-2)^p + (-1)^q \equiv 0^r \pmod{13}$, which we now know means $4^{p/2} \equiv 1 \pmod{13}$, and this only holds when $p/2$ is a multiple of $6$.
Altogether, we now know that $p = 60p'$, $q = 2q' + 1$, and $r = 60r'$, which is actually a bit more than what we need to proceed. It now makes sense to take the equation modulo $7$, because $2013^{6} \equiv 2015^{6} \equiv 1 \pmod{7}$ due to Fermat's little theorem. So we have
$$
    1^{p/6} + 2014^q \equiv 1^{r/6} \pmod{7}
$$
which has no solutions, because it leads to $2014^q \equiv 0 \pmod 7$, but $7$ does not divide $2014$.
Therefore there are no solutions to this equation over the positive integers. We do have the solution $2013^0 + 2014^1 = 2015^1$ if we allow nonnegative integers; above, we excluded such cases when we assumed things like $2013^p \equiv 0 \pmod{61}$ just because $2013$ is divisible by $61$.

Above, I made claims such as "$2^r \equiv 1 \pmod{61}$ only if $r$ is a multiple of $60$" or "there are no solutions to $4^{p/2} + 1 \equiv 0 \pmod{31}$". These are all finite problems; to understand an exponent modulo $m$, you only have to consider exponents of $1, \dots, \phi(m)$, which is $1, \dots, m-1$ when $m$ is prime. Whatever conclusions we make about the exponent repeat modulo $\phi(m)$. 
I just used brute force in Mathematica to do the calculations.
