I am trying to find all integer solutions to the modular congruence $5x^7+x^2+2x \equiv 2\pmod {28}$
I broke it up into 2 cases: that is $\pmod 2,\pmod 7$.
Using the laws of congruence and Fermat's Little Theorem I got:
$5x^7+x^2+2x \equiv x\equiv1\pmod 2$
and
$5x^7+x^2+2x \equiv x^2+7x\pmod 2$
but I am not sure how to use this result I got. Can anyone help please?
HINT
You should break it up into $\pmod 4$ and $\pmod 7$.
That said, you are progressing in nearly the right way.
Notice that $5x^7+x^2+2x \equiv x^2+7x \equiv x^2 \equiv 2 \pmod 7$ from Fermat's Little Theorem. .
Also , $\pmod 4$ can be devided into two cases, when $x$ is odd and when $x$ is even.
If $x$ is even, note that $2x, 5x^7, x^2$ are all divisible by $4$.
If $x$ is odd, note that $x^2+2x+5x^7 \equiv 7x+1 \equiv 2 \pmod 4$.
