Greatest factor of p^6-1

Note that $p^6-1 = (p-1)(p+1)(p^2+p+1)(p^2-p+1)$ . Let $p \ge 11$ be prime:

• Then $p$ is odd, so $p-1,p+1$ are successive even numbers, and hence one is a multiple of $4$ . Thus $p^6-1$ is a multiple of $8$ .
• If $p \equiv 1 \pmod{3}$ then $p-1$ and $p^2+p+1$ are both multiples of $3$ . If $p \equiv 2 \pmod{3}$ then $p+1$ and $p^2-p+1$ are both multiples of $3$ . Thus $p^6-1$ is a multiple of $9$ .
• Since $p$ and $7$ are coprime, $p^6-1$ is a multiple of $7$ .

Thus $p^6-1$ is a multiple of $8 \times 9 \times 7 = 504$ . Since $13^6 - 1 \; = \; 2^3 \times 3^2 \times 7 \times 61 \times 157 \hspace{2cm} 17^6 - 1 \; = \; 2^5 \times 3^3 \times 7 \times 13 \times 307$ we deduce that $\boxed{504}$ is the largest integer that divides all $p^6 - 1$ for primes $p \ge 11$ .

Firstly we can easily find that $n$ must have no prime factor $p \ge 11$ because if there is a prime $p_1 \ge 11$ , then ${ p }_{ 1 }\nmid { p }_{ 1 }^{ 6 }-1$

Next, let $n = { 2 }^{ a }{ 3 }^{ b }{ 5 }^{ c }{ 7 }^{ d }$ that $a , b, c, d$ are non-negative integers. (or whole numbers) Then, we need to find out the value of $a , b, c, d$ .

Case 1, $a$

By Euler's theorem,

${ p }^{ \varphi \left( 8 \right) }={ p }^{ 6 }\equiv 1 \pmod{8} \rightarrow 8|{ p }^{ 6 }-1\rightarrow a\ge 3$

After that we will show that $a = 3$

${ p }^{ 6 }-1=\left( { p }^{ 2 }-1 \right) \left( { p }^{ 4 }+{ p }^{ 2 }+1 \right) \\ \because 2\nmid { p }^{ 4 }+{ p }^{ 2 }+1\\ \therefore 8|{ p }^{ 2 }-1$

When $p = 11$ , ${ p }^{ 2 }-1={ 11 }^{ 2 }-1=120$ , which is not divisible by $16$ . Therefore, $\boxed { a=3 }$

Case 2, $b$

By Euler's theorem,

${ p }^{ \varphi \left( 9 \right) }={ p }^{ 6 }\equiv 1 \pmod{9} \rightarrow 9|{ p }^{ 6 }-1\rightarrow b\ge 2$

After that we will show that $b = 2$

When $p = 29$ ,

$\quad{ p }^{ 6 }-1\\ ={ 29 }^{ 6 }-1\\ \equiv { 2 }^{ 6 }-1\\ \equiv 63\\ \equiv 9 \pmod{27} \\ \therefore 27\nmid { p }^{ 6 }-1 \text{ for some primes } p\ge11$

Therefore, $\boxed { b=2 }$

Case 3, $c$

When $p = 13$ ,

$\quad{ p }^{ 6 }-1\\ ={ 13 }^{ 6 }-1\\ \equiv { 3 }^{ 6 }-1\\ \equiv 728\\ \equiv 3 \pmod{5} \\ \therefore 5\nmid { p }^{ 6 }-1 \text{ for some primes } p\ge11$

Therefore, $\boxed { c=0 }$

Case 4, $d$

By Euler's theorem,

${ p }^{ \varphi \left( 7 \right) }={ p }^{ 6 }\equiv 1 \pmod{7} \rightarrow 7|{ p }^{ 6 }-1\rightarrow d\ge 1$

After that we will show that $d = 1$

When $p = 47$ ,

$\quad{ p }^{ 6 }-1\\ ={ 47 }^{ 6 }-1\\ \equiv { \left( -2 \right) }^{ 6 }-1\\ \equiv 63\\ \equiv 14 \pmod{49} \\ \therefore 49\nmid { p }^{ 6 }-1 \text{ for some primes } p\ge11$

Therefore, $\boxed { d=1 }$

Finally, we can calculate $n$ by substituting $a,b,c,d$ ,

$n = { 2 }^{ a }{ 3 }^{ b }{ 5 }^{ c }{ 7 }^{ d } = { 2 }^{ 3 }{ 3 }^{ 2 }{ 5 }^{ 0 }{ 7 }^{ 1 } = \boxed { 504 }$