$x^{1682}+22x≡1652 (\bmod3599)$. Hello I'm trying to learn the Chinese Remainder Theorem and now I have the problem from an old exam:
$x^{1682}+22x≡1652 (\bmod3599)$. 
Ok, so what makes this problem difficult for me is the combination of $x^{1682}$ and $22x$.
My start to a solution is: The primefactorization of $3599=59\cdot 61$, and according to the CRT have $f: Z_{3599} \to (Z_{59}\times Z_{61})$ but after this it basically stops. I know how to handle the terms individually but not combined. Can someone please help me.
 A: I'll sketch the method I hinted in the comment. Reducing the polynomial mod $59$ and $61$ and using little Fermat with $1682\equiv 0\pmod{58},\ $ $1682\equiv 2\pmod{60}$ we obtain, for $\,x\not\equiv 0\pmod{\!59}$
$\!\bmod\color{#c00}{59}\!:\ 0\equiv x^{1682}+22x-1652\equiv\ 1+22x-0\iff x\equiv -1/22\equiv -3/66\equiv 56/7\equiv\color{#c00} 8$
$\!\bmod 61\!:\ 0\equiv x^{1682}+22x-1652\equiv x^2\!-22x-5\iff x\equiv -9,-13\equiv a\,$ via quad. formula 
$\!\bmod 61\!:\ a\equiv x \equiv\color{#c00}{8+59k}\equiv 8-2k\iff 2k\equiv 8-a.\ $ For the two roots $\,a\,$ above we get:


*

*$a \equiv \ \ {-}9:\,\  2k\equiv 17\equiv 78\iff k\equiv 39\iff x = 8\!+\!59(39\!+\!61j)\equiv 2309\pmod{\!3599}$

*$a \equiv -13:\,\  2k\equiv 21\equiv 82\iff k\equiv 41\iff x = 8\!+\!59(\underbrace{41\!+\!61j}_{\large k})\equiv 2427\pmod{\!3599}$
Also $\,x\equiv 0\pmod{\!59}\,$ is a root. I'll leave that case for you [same as prior $3$ lines but replace $\color{#c00}8\,$ by $0$]. 
Remark $ $ See here for a precise statement of how the roots are generally lifted via CRT.
