Find all $n>1$ such that $\frac{2^n+1}{n^2}$ is an integer.

Find all $n>1$ such that $\dfrac{2^n+1}{n^2}$ is an integer.

I know that $n$ must be odd, then I don't know how to carry on. Please help. Thank you.

Solution 1:

Let's consider $$\frac{2^n+1}{n^k}$$

If $p$ be the smallest prime that divide $n$

Let $$\operatorname{ord}_p2=d,d\mid(p-1,2n)\implies d\mid 2$$ as $p-1<$ all other primes, so it implies

$$ p\mid (2^2-1)\implies p=3.$$

Let $$3^r||n, 2^{2n}\equiv 1{\pmod {3^{kr}}}\implies \phi(3^{kr})|2n$$ as $2$ is a primitive root of $3^s$ for all $s\ge 1$ (as mentioned in Example 8.1 in the Naoki Sato's solution mentioned in Pantelis Damianou's answer).

This implies $$2\cdot 3^{kr-1}|2n \implies kr-1\le r$$

as $(3^{kr-1},\frac{n}{3^{kr}})=1$.

So, $r(k-1)\le 1$.

(1)If $k>2$, there will be no solution.

(2)If $k=2$ then $r=1;$ let $q>p=3$ be next smallest prime that divides $n$. Now $\operatorname{ord}_q2$ must divide $(q-1,2\cdot 3\cdot \frac{n}{3})$

Then $\operatorname{ord}_q2\mid 6$ as $q-1<$ all primes greater than $3\implies (q-1,\frac{n}{3})=1$

So, $q\mid (2^6-1)\implies q=7$, but $2^7+1=129$ is not divisible by $7$.

So, there is no prime$>3$ that satisfies the given condition $\implies n=3$ if $k=2$.

(3)If $k=1$, there is no restriction on $r>0$

Here $\operatorname{ord}_q2$ must divide $(q-1,2\cdot 3^r\cdot \frac{n}{3^r})=(q-1,2\cdot 3^r)$

So, $q-1=2^c3^d$ as $q<$ any other primes, which implies $ (q-1,2\cdot 3^r)=2\cdot 3^{\min(c,r)}$

Programmatically I have observed that $n=3^s$ keeps $\frac{2^n+1}{n}$ an integer, where $s$ is natural number which can be verified as follows:

As $2$ is a primitive root of $3^s$ for all $s\ge 1,$ so, if $n=3^s$, $\operatorname{ord}_{(3^s)}2=\phi(3^s)=2\cdot 3^{s-1}\implies 2^{\frac{\phi(n)}2}\equiv -1\pmod n$

Now, $\frac{\phi(n)}2=3^{s-1}\implies 2^{3^{s-1}}\equiv -1\pmod {3^s}$.

This implies $$(2^{3^{s-1}})^3\equiv -1\implies 2^{3^s}\equiv -1\pmod {3^s}.$$