Do we calculate the power with Chinese theorem?

Question

For a natural numbr $n$ we know that $n=3k_1+1$, $n=4k_2+1$ and $n=5k_3+4$, for natural numbers $k_1, k_, k_3$.

Calculate $5^n\pmod {13}$ .

For that, do we apply the Chinese Theorem?

Answer 1

$n=3k_1+1 \Rightarrow 3k_1+1 \equiv 1 \pmod 4$

$3k_1+1 \equiv 1 \pmod 4 \Rightarrow k_1 \equiv0 \pmod 4$ as a result, $k_1=4m$

$n=5k_3+4 \Rightarrow 12m+1 \equiv 4 \pmod 5$

$12m+1 \equiv 4 \pmod 5 \Rightarrow m \equiv 4 \pmod 5$

$ m \equiv 4 \pmod 5 \Rightarrow m=5k+4$

$m=5k+4 \Rightarrow n=60k+49$

according to Fermat`s Theorem

$5^{12} \equiv 1 \pmod{13}$

$n \equiv 1 \pmod{12} \Rightarrow 5^n \equiv 5 \pmod{13}$

In this case,the key which was about chinese remainder theorem is the fact that we know there exists a solution for n

Answer 2

The fact that $n\equiv 4\pmod5$ is irrelevant.

Since $n\equiv1\pmod3$ and $n\equiv1\pmod 4$, the Chinese Remainder Theorem implies that $n\equiv1\pmod{12}$ is the unique solution mod $12$.

As $5^{12}\equiv1\pmod{13}$ by Fermat's Little Theorem, it follows that $5^n\equiv5\pmod{13}$.

Answer 3

The Chinese remainder theorem will give you a single statement on the modular constraint on $n$, to the modulus $\text{lcm}(3,4,5) = 60$, but the third constraint ($\text{mod } 5$) is not really needed since by Fermat's little theorem we know that Euler's totient $\phi(13) = 13-1 = 12$ sets the periodicity of powers of $a \bmod 13$.

$n\equiv (1,1) \bmod (3,4)\implies n\equiv 1 \bmod 12$ - (constant case CRT)
$5^n \equiv 5^1 \equiv 5 \bmod 13$