Solution: First let us prove that if $b)$ holds $a)$ also holds. $504=7*2^3*3^2$ and we know from Euler's theorem $x^6 \equiv 1(mod7) , (mod8) , (mod9)$ (since $\Phi{(9)}=\Phi{(8)}=6$) So by Chinese remainder theorem we can easily see that if $n$ divides $ 504 $ , $a)$ holds. Secondly let us prove that if $a)$ holds , $b)$ also holds: First assume $5\mid n$. But then for every integer $x$ that is coprime with $n$, $x^6-1$ is divisible by 5. But we know there exist a prime number that is greater than $n$ (so coprime with $n$) and give $2 (mod 5)$ (You can call it from Dirichlet's theorem). If we call that number $p$, $5\nmid p^6-1$ contradiction. Now we know that $(n,5)=1$ so $n$ must divide $5^6-1=15624=504*31$. Suppose $31\mid n$ Since $(11,n)=1$, $31$ must divide $11^6-1=1771560$ but it does not divide so $31 \nmid n $ contradiction. Thus $n \mid 504$ $\blacksquare$