1. **Problem Statement:** Prove that for every integer $a$ and positive integer $n$, the congruence $$a^{13} \equiv a \pmod{2730}$$ holds. 2. **Understanding the modulus:** The modulus 2730 can be factored into primes: $$2730 = 2 \times 3 \times 5 \times 7 \times 13$$ 3. **Using the Chinese Remainder Theorem (CRT):** To prove $$a^{13} \equiv a \pmod{2730}$$ it suffices to prove that $$a^{13} \equiv a \pmod{p}$$ for each prime divisor $p$ of 2730, i.e., for $p = 2, 3, 5, 7, 13$. 4. **Fermat's Little Theorem:** For a prime $p$ and integer $a$ not divisible by $p$, $$a^{p-1} \equiv 1 \pmod{p}$$ which implies $$a^p \equiv a \pmod{p}$$ for all integers $a$ (including those divisible by $p$). 5. **Check each prime modulus:** - For $p=2$, since $2$ is prime, $a^2 \equiv a \pmod{2}$. - For $p=3$, $a^3 \equiv a \pmod{3}$. - For $p=5$, $a^5 \equiv a \pmod{5}$. - For $p=7$, $a^7 \equiv a \pmod{7}$. - For $p=13$, $a^{13} \equiv a \pmod{13}$. 6. **Relate $a^{13}$ to $a$ modulo each prime:** - Since $13 \equiv 1 \pmod{2}$, $a^{13} \equiv a^{1} = a \pmod{2}$. - Since $13 \equiv 1 \pmod{3}$, $a^{13} \equiv a^{1} = a \pmod{3}$. - Since $13 \equiv 3 \pmod{4}$, but more simply, by Fermat's theorem $a^5 \equiv a$, so $a^{13} = a^{5 \cdot 2 + 3} = (a^5)^2 a^3 \equiv a^2 a^3 = a^5 \equiv a \pmod{5}$. - Similarly for $7$, $a^{13} = a^{7 + 6} = a^7 a^6 \equiv a a^6 \pmod{7}$, but since $a^6 \equiv 1$ or $0$, the result holds. - For $13$, directly $a^{13} \equiv a \pmod{13}$. 7. **By the Chinese Remainder Theorem:** Since $a^{13} \equiv a$ modulo each prime factor of 2730, it follows that $$a^{13} \equiv a \pmod{2730}$$ **Final answer:** $$\boxed{a^{13} \equiv a \pmod{2730}}$$