Find two different prime factors of $2^{29}+1$

Question

Given $N=2^{29}+1$, we want to find two different prime factors of $N$. Now, we have $2^{29}+1\equiv (-1)^{29}+1\equiv0\pmod 3$, hence $3\mid N$. Next, we have $$N=2^{29}+1\equiv 2(2^7)^4+1\equiv 2(2\cdot 59+10)^4+1\equiv 2\cdot 10^4+1\pmod {59}$$ $$N\equiv 2(-18)^2+1\equiv 648+1=11\cdot 59\equiv 0\pmod{59},$$ so that $59\mid N$. I already proved that if $p$ is a prime with $p\mid N$, then $p=3$ or $p=58k+1$ for some $k$; that's where the guess to choose $59$ came from. Is there a more elegant way to show that $59\mid N$? I was thinking of using Fermat's little theorem together with the fact that $2^{58}-1=(2^{29}+1)(2^{29}-1)$, but I'm unable to prove that $\gcd(2^{29}-1,59)=1$.

Answer 1

Factorising $$2^{29}+1=3(2^{28}-2^{27}+\cdots +1)$$ reveals that $3$ is one of the prime factor of $2^{29}+1$. As OP said, by the Fermat Little Theorem, $$(2^{29}-1)(2^{29}+1)=2^{58}-1\equiv 0 \quad \pmod {59}$$ with the fact that $(2^{29}-1, 2^{29}+1)= (2, 2^{29}+1) =1$ tells us that $59$ divides either $2^{29}-1$ or $2^{29}+1$.

Using the Euler’s criteria, we have
$$ 2^{29}=2^{\frac{59-1}{2}} \equiv 1 \text { iff } 2 \equiv x^2(\bmod 59) $$ However, the set of quadratic residues $$ \begin{aligned}S&= \left\{n^2(\bmod 59): n \in \{0,1,2,\cdots,58\} \right\} \\&= \{0,1,3,4,5,7,9,12,15,16,17,19,20,21,22,25,26, 27,28,29,35,36,41,45,46,48,49,51,53,57\}\end{aligned}$$ Since $2\not\in S$, therefore $ 2^{29}-1\not \equiv 0 \quad \pmod {59}$ and then $59$ doesn’t divides $2^{29}-1$ and hence $59$ divides $2^{29}+1$. Thus $2^{29}+1$ has two prime factors $3$ and $59$.

Wish it helps!

Answer 2

Is there a more elegant way to show that $59\mid 2^{29}+1$?

By Euler's criterion, $2^{29}\equiv\left(\dfrac2{59}\right)\bmod59$.

There is no need to compute all thirty quadratic residues $\bmod 59$.

We can know that $\left(\dfrac2{59}\right)=-1$ from the Second Supplement to Quadratic Reciprocity;

cf. here and here.