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:

$5x^7+x^2+2x \equiv x\equiv1\pmod 2$


$5x^7+x^2+2x \equiv x^2+7x\pmod 2$

but I am not sure how to use this result I got. Can anyone help please?

  • 1
    $\begingroup$ Stop cheating on your Math 135 assignments! I will quote that on the first page of this assignment it says "Do not look up full or partial solutions on the Internet or in printed sources." $\endgroup$ – user323215 Mar 16 '16 at 0:31


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 $x^2+2x+5x^7 \equiv 7x+1 \equiv 2 \pmod 4$.


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