<s>[INST] How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? My attempt was:
$x^3 + 2x + 2 \equiv 0 \pmod{25}$
By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:
$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 125k^3 + 75k^2 + 25k + 5 \equiv 0 \pmod{25}$$ 
$$\Leftrightarrow 5 \equiv 0 \pmod{25}$$
And I was stuck here :( because k was completely cancelled out, how can we find solution for this equation?  
Thanks,
 [/INST] For this, I think the quickest way is simply to work out the 25 possibilities.  
For 0,1,2,...24 you have $x^3+2x+2$ giving 2, 5, 14, 35, 74, 137, 230, 359, 530, 749, 1022, 1355, 1754, 2225, 2774, 3407, 4130, 4949, 5870, 6899, 8042, 9305, 10694, 12215, 13874 equivalent modulo 25 to 2, 5, 14, 10, 24, 12, 5, 9, 5, 24, 22, 5, 4, 0, 24, 7, 5, 24, 20, 24, 17, 5, 19, 15, 24. 
So the solution is $x \equiv 13$
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