# Show that for all integers congruent modulo 11

## Homework Statement

Let ##a, b, c \in \mathbb{N \setminus \{0 \}}##. Show that for all ##n \in \mathbb{Z}## we have

$$n^{11a + 21b + 31c} \equiv n^{a + b + c} \quad (mod \text{ } 11).$$

## The Attempt at a Solution

We have to show that ##11 | (n^{11a + 21b + 31c} - n^{a + b + c}) \iff \exists k \in \mathbb{Z} : n^{11a + 21b + 31c} - n^{a + b + c} = 11k.##

I tried showing that ##11 | n^{11a + 21b + 31c}## and ##11 | - n^{a + b + c}## and then conclude that ## 11 | (n^{11a + 21b + 31c} - n^{a + b + c})##. I also tried breaking down the variables in even and odd but that gave me too many cases which became tedious very fast. I also tried induction on a keeping b,c fixed.

fresh_42
Mentor
Have you tried to work with ##(n^{a+2b+3c})^{10} \equiv 0 \operatorname{mod} 11## and perhaps the divisibility criterion of ##11## (alternating digit sum)? Or consider which remainders modulo ##11## become ##0## if taken to the ##10-##th power.

What I've done is used Fermat's Little Theorem like this

\begin{equation*}
\begin{split}
n^{11a + 21b + 31c} & \equiv n^{11a}n^{21b}n^{31c} \text{ (mod 11)} \\
& \equiv (n^{a})^{11}(n^{b})^{11} (n^{b})^{10}(n^{c})^{11}(n^{c})^{11}(n^{c})^{9} \text{ (mod 11)} \\
& \equiv n^{a}n^{b} (n^{b})^{10}n^{c}n^{c}(n^{c})^{9} \text{ (mod 11)} \\
& \equiv n^{a}(n^{b})^{11}(n^{c})^{11} \text{ (mod 11)} \\
& \equiv n^{a}n^{b}n^{c} \text{ (mod 11)}.
\end{split}
\end{equation*}

Since 11 is a prime number we have by Fermat's Little Theorem that

\begin{equation*}