What is the largest integer m satisfying $a$$^{12}$ = 1 (mod $m$) for all integers a relatively prime to m?

I believe this can be solved using a homomorphism. So, let $m$ = ${p_1}^{e_1}$...${p_k}^{e_k}$ be the prime factorization of $m$. Then there is a homomorphism from

(Z/$m$Z)* ---> (Z/${p_1}^{e_1}$Z)* x ... x (Z/${p_k}^{e_k}$Z)* which takes [a] ---> (a (mod ${p_1}^{e_1}$), ...., a (mod ${p_k}^{e_k}$))

Where do I go from here?

  • $\begingroup$ Welcome to SE! Could you please edit your text and write it using LaTeX so that it's more readible? People will be much more willing to answer if they can read easily. $\endgroup$ – user88595 May 4 '14 at 17:34


What we need is Carmichael Function $$\lambda(m)=12$$

If prime $p$ divides $m\implies p-1$ must divide $12$

more generally, if $a$ is the highest power of $p$ that divides $m,$ then $\lambda(p^a)$ must divide $\lambda(m)=12$


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