Solutions for $2^n+1=p^q$

Question

I have a problem I can’t solve, please help!

Find all positive integer triples $(n,p,q)$ satisfying $2^n+1=p^q$, where $p,q>1$.

There is a similar problem I can solve: Prove that it is not possible that $2^n-1=p^q$, if $p,q>1$.

My solution: We need to prove that $2^n=p^q+1$ is not possible. Note that $p$ is an odd number and if you check $\mod 4$ then you find that $q$ is also an odd number. Then $2^n=p^q+1=p^q+1^q=(p+1)(p^{q-1}-p^{q-2}+\dots -p+1)$. Note that $2^n$ doesn’t have an odd divisor $>1$, but since $(p^{q-1}-p^{q-2}+\dots -p+1)$ is an odd number $>1$, contradiction.

Answer 1

I try to explain an elementary proof. First, we consider the equation $2^n=p^q-1$. We have $$2^n=p^q-1=(p-1)(p^{q-1}+p^{q-2}+\cdots + p+1)$$ Hence $p$ is odd, $p^{q-1}+p^{q-2}+\cdots + p+1$ is even, and hence $q$ must be even. But if $q=2s$, then $$2^n=p^{2s}-1=(p^s-1)(p^s+1)$$ so $p^s=2^r+1=2^t-1$ for some $r,t$ with $r+s=n$, hence $2^t=2^r+2=2(2^{r-1}+1)$, so $2^{r-1}$ is odd, which only happens when $r=1$, so $t=2$, $n=3$ and hence $p=3$, $q=2$.

The second case the proof you wrote works perfectly.

Answer 2

These are both special cases of Mihăilescu's theorem, which says that the only solution to $$x^a-y^b=1,$$ in integers $a,b>1$ and $x,y>0$ is $(a,b,x,y)=(2,3,3,2)$.

Answer 3

As $p-1\mid p^q-1=2^n$, we see that $p$ must be one more than a power of $2$, so $p=2^k+1$ (with $k\ge1$ because $p=2^0+1$ leads only to $n=0$, $q=1$).

If $q$ is even, then $2^n=p^q-1=(p^{q/2}-1)(p^{q/2}+1)$, so $p^{q/2}\pm1$ must be powers of $2$, making $p^{q/2}=3$, $p=3$, $q=2$. We have found the solution $$ 2^3+1=3^2.$$

If $q$ is odd, then $$2^n=q^p-1=(1+2^k)^q-1 =1+q2^k+(\ldots)2^{2k}-1=\bigl(q+(\ldots)2^k\bigr)\cdot 2^k$$ is a proper odd multiple of $2^k$, contradiction.