Find set of all integers x for which the following holds: $35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod {55}$
Since $55 = 5\cdot 11$, simultaneous congruences:
$35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod 5$
$35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod {11}$
And then these can be simplified:
$3x^{25} - x^{21} \equiv 1 \pmod 5$
$2x^{31} - 3x^{21} \equiv 1 \pmod {11}$
How can I proceed from here?
We will find Fermat's Theorem useful. Recall that it says that if $p$ is prime and $a$ is not divisible by $p$, then $a^{p-1}\equiv 1 \pmod{p}$.
We start as you did. Modulo $5$, we arrive at the congruence $3x^{25}-x^{21}\equiv 1\pmod{5}$. If $x$ is a solution, it cannot be divisible by $5$. Thus $x^4\equiv 1\pmod{5}$ by Fermat's Theorem. Thus $x^{24}=(x^4)^6\equiv 1 \pmod{5}$. Similarly, $x^{20}\equiv 1\pmod{5}$. So our congruence is equivalent to $3x-x\equiv 1\pmod{5}$. Solving $2x\equiv 1\pmod{5}$ is quick: we get $x\equiv 3\pmod{5}$.
For the second congruence, start from your $2x^{31}-3x^{21}\equiv 1\pmod{11}$. Note that $x$ cannot be divisible by $11$. Since $x^{10}\equiv 1\pmod{11}$ (Fermat's Theorem again), we get $x^{31}\equiv x\pmod{11}$ and $x^{20}\equiv 1\pmod{11}$. So our congruence is equivalent to $2x-3x\equiv 1\pmod{11}$, and the solution is $x\equiv -1\pmod{11}$.
So we want to solve $x\equiv 3\pmod{5}$, $x\equiv -1\pmod{11}$. For bigger numbers, we might use the machinery of the Chinese Remainder Theorem. But with these small numbers, it s easiest to look at the numbers that are congruent to $-1$ modulo $11$, and stop when we get a number congruent to $3$ modulo $5$. Start at the smallest positive candidate, $10$. Is it congruent to $3$ modulo $5$? No. Next try $21$ (no), $32$ (no), $43$. Bingo, got it!
There is a unique solution modulo $55$, so the general solution is $x\equiv 43\pmod{55}$, that is, all numbers of the form $55k+43$, where $k$ ranges over the integers, positive, negative, and $0$.