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Determine all positive integers $n^6-1$ divisible by each of $2,3,5,7,13$.

Observe that lcm$(2,3,5,7,13) = 2730$.

Also we have $\phi(13) = 12$ and $\phi(2), \phi(3),\phi(5),\phi(7)$ all of them divides $12$.

Hence for any prime $p$ among $2,3,5,7,13$ we have $$n^{13} \equiv n \mod p.$$

We can say that $p | n^{13} - n$ and hence $2730$ divides $n(n^{12}-1)$ for all $n$.

i.e. $n(n^6-1)(n^6+1)$ is divisible by $2730$ for all integers $n$.

But from here how can I find for which integers $n$ we have $n^6-1 \equiv 0 \mod 2730$?

Need some hints.

Bill Dubuque
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User8976
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1 Answers1

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$2\mid n^6-1$ $\Longleftrightarrow$ $2\not\mid n$.

$3\mid n^6-1$ $\Longleftrightarrow$ $3\not\mid n$.

$5\mid n^6-1$ $\Longleftrightarrow$ $n=1,4\pmod 5$.

$7\mid n^6-1$ $\Longleftrightarrow$ $7\not\mid n$. (This equivalence is immediate from Fermat's little theorem for $p=7$.)

$13\mid n^6-1$ $\Longleftrightarrow$ $n=1,3,4,9,10,12\pmod{13}$.

All equivalences above can be proved by straightforward brute force using modular arithmetics.

Here is a more "theoretical" way to prove the last equivalence. By Fermat's little theorem, $13\mid m^{12}-1$ for all integers $m$ that is not divisible by $13$. Since $m^{12}=(m^2)^6$, $13\mid n^6-1$ iff $n$ is a quadratic residue modulo $13$ not divisible by $13$. Those quadratic residues are represented by $1^2, 2^2, 3^2,4^2,5^2,6^2$, which are $1,4,9,3,12,10$ modulo $13$.

Thanks to the five equivalences above, we know that $n^6-1$ is divisible by each of $2,3,5,7,13$ iff $n$ satisfies all following three conditions.

  • None of $2,3,7$ is a factor of $n$.
  • $n=1,4\pmod 5$.
  • $n=1,3,4,9,10,12\pmod{13}$.

In other words, $n^6-1$ is divisible by each of $2,3,5,7,13$ iff $n$ is congruent to one of the following $144$ numbers modulo $2\cdot3\cdot5\cdot7\cdot13=2730$.

$1$ $29$ $61$ $79$ $101$ $121$ $131$ $139$ $\cdots$ $2599$ $2609$ $2629$ $2651$ $2669$ $2701$ $2729$.

We have been assuming $n$ is an integer. The condition that $n$ is positive can be added easily.

Apass.Jack
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