The congruence you're asking about is $$a {21} \equiv a \mod 15$$

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The congruence you're asking about is $$a^{21} \equiv a \mod 15$$

This can be derived using Euler's theorem, which states that if $$a$$ and $$n$$ are coprime (i.e., their greatest common divisor is 1), then $$a^{\phi(n)} \equiv 1 \mod n$$, where $$\phi(n)$$ is Euler's totient function, which counts the positive integers up to a given integer $$n$$ that are relatively prime to $$n$$.
In this case, $$n = 15$$, and the totient function $$\phi(15) = 8$$ because there are eight numbers less than 15 that are coprime with 15: 1, 2, 4, 7, 8, 11, 13, and 14.
So Euler's theorem gives us $$a^8 \equiv 1 \mod 15$$ for all $$a$$ coprime with 15.
Now we want to show that $$a^{21} \equiv a \mod 15$$. We can write $$21 = 8 \times 2 + 5$$. So we have:
$$a^{21} = a^{8 \times 2 + 5} = (a^8)^2 \cdot a^5 = a^5 \mod 15$$
This is because $$(a^8)^2 \equiv 1^2 = 1 \mod 15$$ by Euler's theorem.
So we have shown that for all $$a$$ coprime with 15, $$a^{21} \equiv a^5 \mod 15$$.
But we want to show that $$a^{21} \equiv a \mod 15$$. This is true if and only if $$a^5 \equiv a \mod 15$$.
This can be shown by considering the cases where $$a = 0,3,6,9,12$$ (where $$a$$ is divisible by 3), and the cases where $$a = 0,5,10$$ (where $$a$$ is divisible by 5). In all these cases, it can be verified that indeed $$a^5 \equiv a \mod 15$$.
For the remaining cases where $$a$$ is coprime with 15 (i.e., when $$gcd(a,15) =1$$), we have from Euler's theorem that $$a^4 = (a^8)^{0.5} \equiv 1^{0.5} =1\mod{15}$$. So in these cases we have:
$$a^5 = a\cdot a^4\equiv a\cdot1=a\mod{15}.$$
So in all cases we have shown that indeed $$a^{21} \equiv a\mod{15}$$. This completes the derivation.