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We have the factorization $561=3*11*17$. Because $\text{gcd}(a,561)=1$, there are integers $x,y$ such that $ax+561y=1$. So $ax=1\mod 561$. Since the gcd is $1$, we have $a\not \in 3\mathbb{Z}, 11\mathbb{Z}, 17\mathbb{Z}$. Therefore, $a^{560}\not \in 3\mathbb{Z}, 11\mathbb{Z}, 17\mathbb{Z}$. So $b^{560}=1\mod 561$. Is my reasoning correct? I have seen a problem that was similar to this that used Fermat's Little Theorem, but that can only be used when the number we mod out by is prime.

cap
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  • The proof is not correct. If it were, we would have $a^{32}\equiv 1\pmod{33}$ if $\gcd(a,33)=1$. But this is not true. We cannot jump from $ax+bm=1$ to $a^{m-1}\equiv 1\pmod{m}$. – André Nicolas Aug 24 '15 at 07:16
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    To solve, work separately modulo $3$, $11$, and $17$. In each case use Fermat's Theorem. – André Nicolas Aug 24 '15 at 07:20
  • If you don't like the "original" I picked, then take a look at this instead. This and related questions have been asked so many times, that it might be better to delete the quesion. Leaving the decision to regular users though. – Jyrki Lahtonen Aug 24 '15 at 07:48

1 Answers1

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HINT:

Use Carmichael function, $\lambda(561)=$ lcm$(2,10,16)=80$

For $(a,561)=1,a^{560}=(a^{80})^7\equiv1^7\pmod{561}$

lab bhattacharjee
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