Show that for all integers congruent modulo 11

[Lelouch]

Homework Statement

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

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

Homework Equations

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]

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.

 

[Lelouch]

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*}
(n^{a})^{11} \equiv n^{a} \text{ (mod 11)} \quad \land \quad
(n^{b})^{11} \equiv n^{a} \text{ (mod 11)} \quad \land \quad (n^{c})^{11} \equiv n^{a} \text{ (mod 11)}.
\end{equation*}

Is this correct?

 

[fresh_42]

Yes, that's fine.