Remainder of dividing $x^{137}+x+1$ by $x+5$

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $\Bbb Z_7$.