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Problem. Solve the equation

$$ 4^\alpha+1=q^\beta $$

for prime number $q$ and positive integers $\alpha,\beta$.

Background. This question arose in the context of the following question:

Find all right triangles such that two of the lengths of its sides are powers of a prime, and the length of the other side is an integer.

Or more specifically,

Find all primes $p,q$ and positive integers $\alpha,\beta,r$, such that $$p^{2\alpha}+r^2=q^{2\beta}.$$

Since this question might be related to Mersenne primes, which is currently an unsolved problem in number theory, here I only search for the most accurate description of the solutions, like "$q$ can be a Mersenne prime" or "$\beta$ must be odd".

My attempt. $4^\alpha+1\equiv2\pmod q$, but Fermat's little theorem tells us $q^\beta\equiv1\pmod3$ whenever $q>3$ and $\beta$ is even. Therefore $\beta$ must be odd.

If $\alpha$ is odd then $5\vert4^\alpha+1$, therefore $q=5$. Assume $\alpha$ is even and $\alpha=2\alpha_1$. If $\alpha_1$ is odd then $q=17$. Similarly we finally get to $q=5,17,11,13,\cdots$.

Moreover we have $r:=\delta_{q^\beta}(4)\vert q^{\beta-1}(q-1)=\varphi(q^\beta)$, which means $q\vert r$ or $r\vert(q-1)$. Here $\delta_p(a)$ denotes the order of $a$ modulo $p$.

Are there any further observations?

atzlt
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  • I'm confused by what you mean by $4^\alpha + 1 \equiv 2 \pmod{q}$ at the beginning, since clearly based on the original equation we would require $4^\alpha + 1 \equiv 0.$ – Stephen Donovan Apr 15 '22 at 16:24
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    That sentence would make sense if it started $4^\alpha + 1 \equiv 2 \pmod 3$ – aschepler Apr 15 '22 at 16:26
  • Catalan's Conjecture implies that $\beta$ must be $1$, so the problem is essentially equivalent to finding Fermat primes greater than $3$ (namely, $(q, \alpha, \beta)=(2^{2^n}+1, 2^{n-1}, 1)$ is a solution to $4^{\alpha}+1=q^{\beta}$ for any Fermat prime $2^{2^n}+1$). The only known such primes are $5,17,257,$ and $65537$. – Geoffrey Trang Apr 15 '22 at 17:14

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