Supposing $a^{n}\cong1(mod\,m)$ and $t \cong r(mod\,n)$. How to prove that $a^{t}\cong a^{r}(mod\,m)$?

Question

I've been looking at this question but I can't figure out any ways to approach it. This is from a first year undergraduate number theory class.

Saw that, and deleted my comment accordingly. For the corrected question, write $t=r+kn$ for some $k$. Then write out the desired congruence. — lulu, Nov 07 '19 at 18:38
What have you tried? Where are you stuck at? Do you think this is true? — Calvin Lin, Nov 07 '19 at 18:38

score 0 · Answer 1 · answered Nov 07 '19 at 18:40

0

$$t=kn+r$$

$$a^t= a^{kn+r} = (a^n)^k .a^r \equiv 1.a^r= a^r, \mod (m)$$

answered Nov 07 '19 at 18:40

Mohammad Riazi-Kermani

68,728

score 0 · Accepted Answer · answered Nov 07 '19 at 18:41

0

Since $t \cong r \mod n$, then there is some integer $q$ such that $t-r=nq$. This gives us $a^{t-r}=a^{nq}=\left(a^n\right)^q=1^q=1 \mod m$. This give $a^t \cong a^r \mod m$.

answered Nov 07 '19 at 18:41

David Pement

836

Supposing $a^{n}\cong1(mod\,m)$ and $t \cong r(mod\,n)$. How to prove that $a^{t}\cong a^{r}(mod\,m)$?

2 Answers2