# Product of all integers less than the largest 12 digit prime number

For a prime number $p$ the set of co-primes less than or equal to it is given by $\{1,2,3,4,...p-1\}$. For $0<x<p$, we define $f(x,p)= 1$ if and only if all the numbers from $1$ to $p-1$ can be written as a power of $x$ in modulo-$p$ arithmetic. Let $n$ be the largest $12$-digit prime number . Find the product of all integers $j$ less than $n$ such that $f(j,n)=1$.

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More editing will be required to make the definition of $f$ coherent, but I've made it so that the text no longer disappears by adding dollar signs. – Jonas Meyer Jan 27 '12 at 17:14
After a mistaken reading, I think I understand the question. We have $f(x,p)=1$ iff $x$ is a primitive root of $p$. Let our big prime be $p$. You want to find the product of the $\varphi(p-1)$ primitive roots of $p$. This product is easily congruent to $1$ modulo $p$, but that does not help much in finding the exact value. – André Nicolas Jan 27 '12 at 17:40
There are very interesting questions here. Define $\Lambda(n)$ as the product of primitive roots modulo $n$ lifted up to $\mathbb{Z}$ premultiplication. $\bf(a)$ It's only defined for $n=1,2,3,p^r,2p^r$ for odd primes $p$ (when $Z_n^\times$ is cyclic IOW); can it be extended to other $n$ in a natural way, like Legendre to Jacobi symbols for instance? $\bf(b)$ What is the relation between $\Lambda(2p^r)$ and $\Lambda(p^r)$, or $\Lambda(p^r)$ to $\Lambda(p^s)$ with $r>s$? (cont'd) – anon Jan 27 '12 at 19:52
... $\bf(c)$ Does this construction obtain any sort of convergence or induce any particularly special function in the $p$-adic realm? $\bf(d)$ What does $\Lambda(n)$ look like asymptotically? Or the valuation part $l(p^r)=v_p(\Lambda(p^r)-1)$? Or the non-$p$ part given by $(\Lambda(p^r)-1)p^{-l(p^r)}$? – anon Jan 27 '12 at 19:52
(that should be a 4, not a 3, above) – anon Jan 27 '12 at 20:05