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In $\mathbb{Z}_7[x]$, what is the remainder of dividing $x^{137}+x+1$ by $x+5$?

I can not find how to solve this problem of modular arithmetic. Anybody could tell me only as I proceed to solve this exercise?

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    $\begingroup$ Do you think you can simplify $x + 5$ in $\mathbf{Z}_5[x]$? $\endgroup$
    – Dave
    Sep 11, 2014 at 23:06
  • $\begingroup$ no. It is a mistake. is In $\mathbb{Z}_7[x]$, $\endgroup$
    – Jose
    Sep 11, 2014 at 23:13
  • $\begingroup$ Do you know about quotient rings? $\endgroup$
    – Dave
    Sep 11, 2014 at 23:15
  • $\begingroup$ little bit. I'm trying to solve a problem that said, what is the remainder of dividing x ^ 137 + x + 1 in x + 5? in $\mathbb{Z}_7[x]$ $\endgroup$
    – Jose
    Sep 11, 2014 at 23:20
  • $\begingroup$ Yes, I understand. The reason I was asking is because quotient rings provide a way to understand why the remainder theorem works. Matt Rigby's answer below is correct. $\endgroup$
    – Dave
    Sep 11, 2014 at 23:22

3 Answers 3

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The remainder theorem still works, so it's equal to $(-5)^{137} + (-5) + 1 \mod 7$.

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Fermat's Little Theorem says that, modulo $7$, $x^7 \equiv x$ for all $x \in \mathbb{Z}$. That provides you a simple method for reducing powers: $x^{137} \equiv x^{131} \equiv x^{125} \equiv \cdots x^{5}$. That reduces it to a simpler problem. Combine this with Matt Rigby's solution and the problem should be pretty easy to finish.

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Over any ring, the polynomial $x-a$ divides $x^n-a^n$ for all natural numbers $n$.Then, for our case, we will have $$x+5\ \mid\ \left(x^{137}-(-5)^{137}\ +\ x-(-5)\ +\ 1-1\right) $$ Using the congruence notation, this is $$x^{137} + x + 1 \equiv (-5)^{137}+(-5)+1 \pmod{x+5}\,.$$ Note that we indeed expect a constant polynomial as the remainder, since it must have strictly less degree than the divisor.

Observe also that $-5$ can be replaced to $2$ as we work over $\Bbb Z_7$.

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