$ n^6 - 1 = (n - 1)(n + 1)(n^2 + n + 1)(n^2 - n + 1)$. $ n^3 - 1 = (n - 1)(n^2 + n + 1)$. $ n^2 - 1 = (n - 1)(n + 1)$. If $ p|n^2 - n + 1$ then $ p|(n - 1)^2.(n^2 + n + 1)(n + 1)$. *)$ p|n - 1$ so $ p|1$, impossible. *)$ p|n^2 + n + 1$, impossible. *)$ p|n + 1$ so $ p = 3$. $ \Rightarrow n^2 - n + 1 = 3^l$. $ \Rightarrow (2n - 1)^2 = 3(3^{l - 1} - 1)$. $ \Rightarrow l = 0, n = 1$, contradiction.

your solution is very nice,but we have $ n^2 + 1 - n = 3^x.r$ and ...

Thjch Ph4 Trjnh wrote: $ n^6 - 1 = (n - 1)(n + 1)(n^2 + n + 1)(n^2 - n + 1)$. $ n^3 - 1 = (n - 1)(n^2 + n + 1)$. $ n^2 - 1 = (n - 1)(n + 1)$. If $ p|n^2 - n + 1$ then $ p|(n - 1)^2.(n^2 + n + 1)(n + 1)$. *)$ p|n - 1$ so $ p|1$, impossible. *)$ p|n^2 + n + 1$, impossible. *)$ p|n + 1$ so $ p = 3$. $ \Rightarrow n^2 - n + 1 = 3^l$. $ \Rightarrow (2n - 1)^2 = 3(3^{l - 1} - 1)$. $ \Rightarrow l = 0, n = 1$, contradiction. WOW you just proved that n=2 is not an answer while it clearly is. the last line of your solution has a flaw. when you 4x the LHS you should 4x the RHS too. EDIT: just use modulo 9 for the last part.

We claim that the only solution is $n=1,2$ It is easy to see that $n=1,2$ satisfies the condition. We now show that that for all $n^6-1$ has more distinct prime factors than $(n^2-1)(n^3-1)$ $\forall n\geq 3$ Note that $n^6-1=(n-1)(n^5+n^4+n^3+n^2+n+1)=(n-1)(n+1)(n^2-n+1)(n^2+n+1)$ and $(n^2-1)(n^3-1)=(n-1)^2(n+1)(n^2+n+1)$ We see that, $gcd(n^2-n+1,n-1)=gcd(n(n-1)+1,n-1)=1$ $\quad \quad \quad$ $gcd(n^2-n+1,n^2+n+1)=gcd(n^2+n+1-2n,n^2+n+1)=gcd(n^2+n+1,2n)=gcd(n(n+1)+1,2)=1$ and lastly, $gcd(n^2-n+1,n+1)=gcd(n(n+1)+1-2n,n+1)=gcd(3,n+1)=1 \;\text{or}\; 3$ now for $n\geq 3 \quad \quad n^2-n+1\geq 3$, so for $n\geq 3$ $n^6-1$ has more distinct prime factors than $(n^2-1)(n^3-1)$. So the only solutions are $n=1,2$.$\blacksquare$

This problem can be solved in this way we clearly see that n=2 is the only solution now write $n^6-1=(n+1)(n^2-n+1)(n^3-1)$ now assume n>2 then we have gcd(n^3-1, n^3+1) <=2 then applying similar property of gcd we obtain that 3| n^2-n+1 now we assume $n^2-n+1=3^x$ for some x element of N therefore then we have on testing n^2-n+1 mod9 for values 2,3,5 we have remainder 3 but this is a contradiction since we assume that n^2-n+1 is a power of 3

It is trivial with Zsigmondy. If $a\neq2$, there is a prime number $p$ that divides $a^n-1$ but don’t divide any $a^k-1$ for any positive integer $k<n$. Since every divisor of $n^6-1$ also divides $(n^2-1)(n^3-1)$, n must be equal to 2. $\blacksquare$

Note: Only exceptions are $n=1$ , $n=2$ and $a+b=2^k$ , $n=6$ and $a=2$

first we will prove that $n^{6}-1>(n^3-1)(n^2-1)$ , clearly this is true as $(n^3-1)(n+1)(n^2+2n+2)>0$ but for $n>1$ this is true so done. Now clearly as $n^{6}-1>(n^3-1)(n^2-1)$ we must have $(n^3-1)(n^2-1)|(n^{6}-1)$ as any prime divisior of $(n^{6}-1)$ is a divisor of $(n^{3}-1)(n^{2}-1)$ in turn , we have $\frac{n^{3}+1}{n^{2}-1} \in \mathbb{Z} \implies \frac{n(n^2-1)+n+1}{n^2-1} \in \mathbb{Z} \implies \frac{n+1}{n^2-1} \in \mathbb{Z} \implies \frac{1}{n-1} \in \mathbb{Z} \implies n=2$ for $n>1$, henceonly $\boxed{n=2}$ works , which on plugging works. $\blacksquare$