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How can I determine all prime factor of $\frac{3^{41} -1}{2}$ without doing prime factorizations.

Any help will be appreciated.

Surojit
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    In a fairly recent question (yesterday?) I gave a proof that $83$ divides our number. The other factors are not easy to get at by hand. – André Nicolas Apr 19 '16 at 17:02
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    Found it! – André Nicolas Apr 19 '16 at 17:04
  • I suppose you can repeatedly use $(a^2-1)=(a-1)(a+1)$ and that $3^n+1$ is even? – jim Apr 19 '16 at 17:06
  • You know that $3$ is a square modulo any prime divisor $p$, so $p\equiv \pm 1\pmod{12}$. You also know that $41\mid p-1$, or $p\equiv 1\pmod {41}$. So $p\equiv 1,83\pmod{41\cdot 12}$. That reduces your search for prime divisors somewhat, but given the actual factorization, probably not enough. – Thomas Andrews Apr 19 '16 at 17:19
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    I'd say that the method is similar as for the Mersenne numbers $2^p-1$ (with $p$ prime) which if composite have divisors of the form $2kp+1$ (in particular the least prime divisor of $2^p-1$ is of this form) – reuns Apr 19 '16 at 17:21
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    The fact that $83$ divides $3^{41}$ can be established by hand: Modulo $83=3^4+2$, we have $3^n\equiv 4\cdot 3^{n-8}$, hence $$3^{41}\equiv 4\cdot 3^{33}\equiv \ldots \equiv 4^{5}\cdot 3\equiv 28\cdot 3\equiv 1$$ – Hagen von Eitzen Apr 19 '16 at 17:24

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Due to the particular form of that number, its prime factors are restricted to a very specific subset of the prime numbers. If $p>41$ and $$ 3^{41}\equiv 1\pmod{p},$$ it follows that the order of $3$ in $\mathbb{F}_p^*$ is exactly $41$, so $41\mid (p-1)$ by Lagrange's theorem, or: $$ p\equiv 1\pmod{41}.$$ The first prime in such a subset is $p=83$. In such a case, $$ 3^{41}\equiv 3^{\frac{p-1}{2}} \equiv \left(\frac{3}{p}\right) \equiv 1\pmod{p}$$ by quadratic reciprocity, since $83\equiv -1\pmod{4}$ gives $\left(\frac{-1}{83}\right)=-1$ and $89\equiv -1\pmod{3}$ gives $\left(\frac{-3}{83}\right)=-1$, so $\left(\frac{3}{83}\right)=1$. You may check that no prime in the interval $[2,41]$ is a divisor of $\frac{3^{41}-1}{2}$, hence $\frac{3^{41}-1}{2}$ is $83$ times the product of some (huge) primes $\equiv 1\pmod{41}$.

It won't be an easy problem in the general case. For instance, to test numbers of the form $2^p-1$ (or $\frac{3^p-1}{2}$ with $p\equiv\pm 5\pmod{12}$) for primality is actually the main way to generate colossal primes (like $\frac{3^7-1}{2}=1093$, that is not so huge but is a neat example). On the other hand, the same argument as above shows that for any prime $p$, there are infinite primes $\equiv 1\pmod{p}$.

Jack D'Aurizio
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