can it be proven that $4^n \mod 15$ is equal to 1 for even n, 4 for odd n?

Question

Can it be proven that $4^n \mod 15$ is equal to 1 if $n$ is even, 4 if $n$ is odd? it seems like this holds for fairly high $n$'s, but I was unable to rigorously prove this statement.

I tried to state that for even $n$, $4^n \mod15=(16)^{\frac{n}{2}}\mod 15=16\mod15=1$, and likewise for odd $n$, $4^n \mod15=4\times((16)^{\frac{n-1}{2}}\mod 15)=4\times(16\mod15)=4$, but I feel as though the logic behind the last equality in each of these statements is faulty.

Is the statement in the title to this question correct and if so how can it be proven, or is it completely wrong and only by chance it seems as though this is the case?

$n\div 2\Rightarrow n = 2q!+!r,\ r\in{0,1},$ so $\bmod \color{#c00}{15}!:\ 4^n! = 4^{r+2q}\equiv 4^r (\color{#c00}{4^2})^q\equiv 4^r \color{#c00}1^q$ is $,4^0 = 1,$ if $,r=0\ (n$ even), and $4^1,$ if $,r=1\ (n$ odd), by congruence power & product rules in linked dupe. — Bill Dubuque, Jun 08 '22 at 14:33

score 1 · Answer 1 · answered Jun 08 '22 at 14:23

1

Generally, your procedure is right. But it is more precise to write as $$4^{2n} \equiv (16)^n \equiv 1^n \equiv 1 (\text{mod } 15).$$ Also, you can write $$4^{2n+1}\equiv(16)^n\cdot 4\equiv1^n\cdot 4\equiv4(\text{mod }15).$$

answered Jun 08 '22 at 14:23

Riemann

1,173

can it be proven that $4^n \mod 15$ is equal to 1 for even n, 4 for odd n?

1 Answers1