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Exercise 2122Number theory › Divisibility and gcdM1Added 3 Nov 2026

Carmichael numbers (statement)

#Carmichael#primes#Fermat

Statement

Define a Carmichael number. Show that 561561 is a Carmichael number.

Hint

561=3⋅11⋅17561=3\cdot11\cdot17. Show a560≡1(mod561)a^{560}\equiv1\pmod{561} for every aa coprime to 561561, by checking mod 3,11,173,11,17 and using CRT.

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Solution

Definition. A composite nn such that an−1≡1(modn)a^{n-1}\equiv1\pmod n for all aa coprime to nn. (Fermat's little theorem fails to detect compositeness of Carmichael numbers.)

561 = 3·11·17. For aa coprime to 561561:
- mod 33: a2≡1a^2\equiv1, so a560=(a2)280≡1a^{560}=(a^2)^{280}\equiv1.
- mod 1111: a10≡1a^{10}\equiv1, so a560=(a10)56≡1a^{560}=(a^{10})^{56}\equiv1.
- mod 1717: a16≡1a^{16}\equiv1, so a560=(a16)35≡1a^{560}=(a^{16})^{35}\equiv1.

CRT. Since 3,11,173,11,17 are pairwise coprime and each divides a560−1a^{560}-1, their product 561561 divides a560−1a^{560}-1. So a560≡1(mod561)a^{560}\equiv1\pmod{561}.

Conclusion. 561561 is a Carmichael number.