I am looking for a proof of the equality in the title (where $q\in\mathbb N$, ($q\ge2$)). Does anyone know such a proof?
Thanks in advance
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19$\begingroup$ Since $q^n-1=\prod_{d\mid n} \Phi_d(q)$, where $\Phi_d$ is the $d$th cyclotomic polynomial, we have $lcm(q-1,\dots,q^n-1) = lcm(\Phi_1(q),\dots,\Phi_n(q))$. But it's easy to see via the Euclidean algorithm that $\gcd(q^i-1,q^j-1)=q^{\gcd(i,j)}-1$, which implies that the $\Phi_k(q)$ are almost totally relatively prime with one another. So the quantity in question is almost exactly $\prod_{k\le n} \Phi_k(q) \approx \prod_{k\le n} q^{\phi(k)} = q^{\sum_{k\le n} \phi(k)}$. That's where the $3n^2/\pi^2$ comes from. $\endgroup$– Greg MartinJun 1, 2014 at 7:57
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1 Answer
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Niven, Fermat's theorem for matrices, Duke Math. J. 15 (1948) 823–826, MR0026672 (10,183e), proved, for any positive integers $a$ and $r$, $${\rm lcm}[x-1,x^2-1,\dots,x^r-1]_{x=a}={\rm lcm}[a-1,a^2-1,\dots,a^r-1]$$ Now follow Greg Martin's comment.
answered Jun 2, 2014 at 0:24