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,