I need to show that if $a,b,k$ and $m$ are integers and $k ≥ 1, m ≥ 2$, and $a ≡ b\pmod m$, then: $a^k ≡ b^k \pmod m$.

But I have no idea how to show this, I have never been this confused. So is there anyone who could help? just a little, like I honestly feel overwhelmed (sounds stupid I know, sorry)

*what do i need to do with the m ≥ 2 ???


I assume you know (all equivalences are $\text{mod } m$)

  1. $a\equiv b \iff a-b\equiv 0$
  2. $c\equiv 0 \implies cd\equiv 0$


$\begin{align}&a\equiv b\\ \iff &a-b\equiv 0\\ \implies &(a-b)(a^{k-1} + a^{k-2}b + \cdots + ab^{k-2}+b^{k-1})\equiv 0\\ \iff &a^k-b^k\equiv 0\\ \end{align}$

The last deduction above is technically hand-waved but can be made formal with a summation or with induction.


Hint: For $k≥1$,

$a^k-b^k$ is divisible by $a-b$.

That is, $a≡b \pmod m$ implies $m$ divides $a-b$.

And also $a-b$ divides $a^k-b^k$, thus by transitivity, $m$ divides $a^k-b^k$.

(i.e) $a^k ≡ b^k\pmod m$.

Note: $(a-b)(a^{k-1} + a^{k-2}b + \cdots + ab^{k-2}+b^{k-1})=a^k-b^k$


$a \equiv b \pmod m$ means $a = b+vm$ for some integer $v$.

So $a^k = (b + vm)^k = \sum_{j=0}^k {k\choose j}(vm)^jb^{k-j}$ by the binomial theorem.

The trick is that for each $j \ge 1$ that ${k\choose j}(vm)^jb^j$ is a multiple of $m$.

So $(b + v*m)^k = b^k + \sum_{j=1}^k {k\choose j}(vm)^jb^{k-j}\equiv b^k \pmod m$

So $a^k = (b + v*m)^k \equiv b^k \pmod m$

(If you really care $a^k = b^k + m*K$ where $K = \sum_{j=1}^k {k\choose j}v^jm^{j-1}b^{k-j}$. )


Actually I find it easier to just do the product rule.

If $a \equiv c\pmod m$ and $b \equiv d\pmod m$ then $ab \equiv cd \pmod m$. Because there are integers $k,j$ so that $a = c + km$ and $b = d + jm$.

So $ab = cd + cjm + dkm + jkm^2 = cd + m(cj + dk + jkm)$.

So $ab \equiv cd \pmod m$.

So by induction we know that if $a \equiv b \pmod m$ then $a^2 = a\cdot a \equiv b\cdot b\pmod m = b^2$ and via induction $a^k = a^{k-1}\cdot a \equiv b^{k-1}\cdot b \pmod k= b^k$


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