Least residue power modulo

Question

May someone confirm that the least residue of $44^8$ modulo $7$ is $4$ please?

And that $2^2 \equiv 4\, (\!\!\!\mod 7)$?

Thanks in advance.

$44^6\equiv1\pmod7$ by Fermat's little theorem, so $44^8\equiv44^2\equiv2^2\equiv4\pmod7$ — J. W. Tanner, Jan 07 '21 at 22:25

Bernard · Answer 1 · 2021-01-07T22:21:34.993

2

It is correct: $44\equiv 2\mod 7$, so $44^8\equiv 2^8\mod 7$. Now $2^3\equiv 1\mod 7$, so $2^8\equiv 2^{8\bmod 3}=2^2\mod 7=4$, since $0\le 4<7$.

edited Jan 07 '21 at 22:21

answered Jan 07 '21 at 22:08

Bernard

Thanks for your help However may you kindly explain the part "2^2 = 1 mod 7" please – Joann Sammut Jan 07 '21 at 22:17
Which part should be explained? – Bernard Jan 07 '21 at 22:18
The part 2^2 = 1 mod 7 – Joann Sammut Jan 07 '21 at 22:20
1

Oh! Sorry, it was a typo, and I forgot to check before posting. I meant $2^3$ (which explains the $2^{8\bmod 3}$). It's fixed now. – Bernard Jan 07 '21 at 22:23
1

Thanks once again – Joann Sammut Jan 07 '21 at 22:30

score 0 · Answer 2 · answered Jan 07 '21 at 23:11

Lemma needed : For $a,b,n,r \in \mathbb{Z^+},$

$$a \equiv b \pmod{n} \implies a^r \equiv b^r \pmod{n}.$$

Proof:

$\displaystyle a\equiv b \pmod{n} \implies n | (a - b) \implies$

$\displaystyle \exists ~k \in \mathbb{Z} ~\text{such that}~ a = (b + nk) \implies$

$\displaystyle (a^r - b^r) = (b + nk)^r - b^r$

[Using binomial expansion]
$\displaystyle =~ \sum_{i=0}^r \left[(b)^i (nk)^{r - i}\right] ~-~ b^r$

$\displaystyle =~ \sum_{i=0}^{(r-1)} \left[(b)^i (nk)^{r - i}\right].$

Since the above summation stops at $(r-1)$, $n$ divides each term in the summation.

Therefore
$n | (a^r - b^r) \implies a^r \equiv b^r \pmod{n}.$

2 Answers2