How to solve congruence like these?

Question

So i have three weird congruence that i don't know how to start:

I know that i have to use euler-fermat formula, but i don't know how to start. All i did was to check a and m are relative prime numbers.

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Sep 28 '23 at 14:27
You'll want Euler totient's function. $a^{\phi(n)} \equiv 1 \pmod {n}$ and probably iterate it — TurlocTheRed, Sep 28 '23 at 14:31
@TurlocTheRed - However, (3) looks like a trick question. Also, Carmichael's function is generally more useful than Euler's in this kind of problem. — mr_e_man, Sep 28 '23 at 17:49
Apply mod order reduction as in the dupe. Hints: (with $\lambda$ = Carmichael lambda) $\begin{align} \ &(1)\ \ \lambda(600=8\cdot 3\cdot 25)={\rm lcm}(4,2,20)=20,\ \ {\rm so}\ \ 2021^{1+20k} \equiv 2021!!!\pmod{600}\[.2em] &(2)\ \bmod \phi(25)!=!20!:\ 47^{4k}\equiv 1\ {\rm by}\ \lambda(4\cdot 5)={\rm lcm}(2,4)=4;\ {\rm or\ directly}\ 7^4\equiv 1\[.2em] &(3)\ 13^{n+1} \bmod 26 = 13(13^n \bmod 2)=13 \end{align}$ $\tag*{}$ — Bill Dubuque, Sep 28 '23 at 18:51
@BillDubuque $(1)$ - Actually $\lambda(8)=2$, not $4$. But the $20$ in LCM makes it irrelevant. — mr_e_man, Sep 28 '23 at 19:25
@mr_e_man Yes, of course. Originally I wrote it using Euler $\phi$ but forgot to update it, but it has no effect (even if it were a proper multiple of $\lambda$ we can still use it for exponent reduction). — Bill Dubuque, Sep 28 '23 at 19:45

0 Answers0