What is the remainder of $(2012^{2013}+2013^{2012}):(2012\times 2013)$? What is the remainder if the sum $$(2012^{2013}+2013^{2012})$$
is divided by $$2012\times 2013$$

We don't need Euler Totient Theorem
As $a+1\equiv1 \pmod a,$
$$N=a^{a+1} + (a+1)^a\equiv (a+1)^a\pmod a\equiv1^a \equiv a$$
$\implies N=1+a\cdot A$ for some integer $A$ 
Similarly, as $a\equiv-1 \pmod{a+1},$
$$N=a^{a+1} +(a+1)^a\equiv a^{a+1}\pmod{a+1}\equiv(-1)^{a+1}\equiv 
\begin{cases}
   1 & \text{ if } a \text{ is odd} \\
   -1 &   \text{ if } a \text{ is even}
  \end{cases}$$
If $a$ is odd, $N\equiv1\pmod {a+1}\implies N=1+(a+1)B$  for some integer $B$
Then $1+a\cdot A=N=1+(a+1)B\implies \frac{(a+1)B}a=A$ which is an integer
$\implies a$ divides $(a+1)B$
$\implies a$ divides $B$ as $(a+1,a)=1$
So, $B=a\cdot C$ where $C$ is some integer
Then $N=1+(a+1)B=1+(a+1)\cdot a\cdot C\equiv 1\pmod{a(a+1)}$
If $a$ is even, $1+a\cdot A=N=-1+(a+1)B\implies (a+1)B-a\cdot A=2=2(a+1-a)$
$\implies \frac{a(A-2)}{a+1}=B-2$ which is an integer
$\implies (a+1)$ divides $a(A-2)$
$\implies (a+1)$ divides $(A-2)$ as $(a,a+1)=1$
So, $A=2+(a+1)D$  for some integer $D$ 
So, $N=1+a\cdot A=1+a\{2+(a+1)D\}=2a+1\pmod{a(a+1)}$
Here $a=2012$ which is even, so the remainder will be $2a+1=2\cdot2012+1=4025$ 
