TITLE: Question regarding solving polynomial of congruence?
QUESTION [1 upvotes]: In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?  
Thanks,

REPLY [1 votes]: HINT $\rm\ \ x\ p(x)\ =\ 2\ x^4 + 7\ x^2 - 4\ x\ \equiv\ 2\ (x-1)^2\ \ (mod\ 5)\ $ for $\rm\ x\not\equiv 0\ $ since then $\rm\ x^4\equiv 1\ $
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