<s>[INST] Remainders problem What will be the reminder if $23^{23}+ 15^{23}$ is divided by $19$?
 Someone did this way: 

$15/19 = -4$ remainder and $23/19 = 4$ remainder So $(-4^{23}) + (4^{23}) =0$ but i didn't understand it

 [/INST] We have $23\equiv 4\pmod{19}$ and $15\equiv -4\pmod{19}$. Then for any odd positive integer $n$, we have $23^n\equiv 4^n \pmod{19}$ and $15^n\equiv (-4)^n \equiv (-1)^n4^n\equiv -(4^n)\pmod{19}$. 
Add. We get $4^n -4^n\equiv 0 \pmod{19}$.
Added: We don't really need to go to $4$ and $-4$. For note that $23\equiv -15 \pmod{19}$. So if $n$ is any odd integer, then $23^n +15^n\equiv (-15)^n +15^n\equiv -15^n+15^n\equiv 0\pmod{19}$.
And we do not really need congruence notation. For recall that if $n$ is odd, then $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots +b^{n-1}$. Let $a=23$ and $b=15$. We get that $23+15$  divides $23^{23}+15^{23}$. This yields the desired result.
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