If $p$ and $q$ are coprime positive integers s.t. $\frac{p}{q}=\sum_{k=0}^{100}\frac1{3^{2^k}+1}$, what is the smallest prime factor of $p$?

Question

If the sum $$S=\frac14+\frac1{10}+\frac1{82}+\frac1{6562}+\cdots+\frac1{3^{2^{100}}+1}$$ is expressed in the form $\frac pq,$ where $p,q\in\mathbb N$ and $\gcd(p, q) =1.$ Then what is smallest prime factor of $p$ ?

We have: $$S=\sum_{k=0}^{100} \frac1{3^{2^k}+1}.$$ Please give me some hints to evaluate such kind of sums, in general.

Remark.

• Since $4$ divides the denominator of the zeroth term ($\frac14$) of $S$, but does not divide the denominator of any other term, we can see that $2\nmid p$.

• Note that each term of $S$ is $\equiv 1\pmod{3}$. Therefore, $S\equiv 101\cdot 1\equiv 2\pmod{3}$, so $3\nmid p$.

• Since $5$ divides the denominator of the first term ($\frac1{10}$) of $S$, but not the denominator of any other term, we conclude that $5\nmid p$.

Answer 1

I'm not sure what kind of problem this is.

It seems that the problem is not intended to be done by hand.

If we denote by $S(n)$ the number $\displaystyle\sum_{k = 0}^n \frac 1{3^{2^k} + 1}$, then the first several values of the numerator of $S(n)$ look like this:

1
7
3^3 * 11
974867
20982415713197
3 * 6480139987906036648979676749
13 * 25220504737903 * 1202418613506277 * 84660948985522106511557529679
149 * 883 * 126001 * 11868766710884224982021663692780373317124689104200960317897970407656906279023556512105818421377935790975902821

Of course, it is easy to show that $3$ divides the numerator if $n \equiv 2 \pmod 3$. This would be a much more reasonable exercise in elementary number theory.

However we have here $n = 100$. This leads to something without a pattern.

With the help of some computer algebra system, I am able to find that the smallest prime factor of the numerator is $37$. This is done by checking all primes up to $37$ one by one. So it is not really doable with paper and pencil.