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DavyDuf wrote: Fake solve (?) From the problem we know that for $n \equiv 8 mod 337 \ \ \rightarrow \ \ n^7 \equiv 1 \mod 337$ $n^7 \equiv 1 \mod 3 \ \ \rightarrow \ \ n \equiv 1 \mod 3$ $n^7 \equiv 1 \mod 2 \ \ \rightarrow \ \ n \equiv 1 \mod 2$ Quick CRT, you will get the smallest number of $n$ which is 1356 A quick bruteforce search shows that the minimal value of $n$ is $79$.

$8^7\equiv 1\pmod{337}\implies (8^k)^7\equiv 1\pmod{337}$ By brute force, we have $$8^1,8^2,8^3,8^4,8^5,8^6,8^7\equiv 8,64,175,52,79,295,1\pmod{337}.$$These are the roots of $n^7-1\equiv 0\pmod{337}$. We also need $n\equiv 1\pmod {6}$ for $6$ to divide $n^7-1$ so $\boxed{79}$ is the smallest number. (since $1$ isn’t allowed.)

Quidditch wrote: $8^7\equiv 1\pmod{337}\implies (8^k)^7\equiv 1\pmod{337}$ By brute force, we have $$8^1,8^2,8^3,8^4,8^5,8^6,8^7\equiv 8,64,175,52,79,295,1\pmod{337}.$$These are the roots of $n^7-1\equiv 0\pmod{337}$. We also need $n\equiv 1\pmod {6}$ for $6$ to divide $n^7-1$ so $\boxed{79}$ is the smallest number. (since $1$ isn’t allowed.) Could there exist other residues not of the form $8^k$?

TYT wrote: Quidditch wrote: $8^7\equiv 1\pmod{337}\implies (8^k)^7\equiv 1\pmod{337}$ By brute force, we have $$8^1,8^2,8^3,8^4,8^5,8^6,8^7\equiv 8,64,175,52,79,295,1\pmod{337}.$$These are the roots of $n^7-1\equiv 0\pmod{337}$. We also need $n\equiv 1\pmod {6}$ for $6$ to divide $n^7-1$ so $\boxed{79}$ is the smallest number. (since $1$ isn’t allowed.) Could there exist other residues not of the form $8^k$? There should be at most $7$ roots of the equation $n^7-1\equiv 0\pmod{337}$ so there is no other root.