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Consider a prime power $q$ having the form $16t^2+1$, where $t$ is a positive integer. Numerical experiments show that when $t \leq 10^9$, each prime power $q$ with this form is indeed a prime.

In general, is it true that a prime power of this form must be a prime?

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  • $\begingroup$ I think Granville has a result that shows r is either 1 or less than 6 for q^r -1 a square. See arxiv.org/abs/1212.6306 for his result and applicability to your problem. Gerhard "At Least It's A Start" Paseman, 2017.02.16. $\endgroup$ Commented Feb 16, 2017 at 20:26
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    $\begingroup$ Thank you so much. With the referene you mentioned, I found a comprehensive treatment on this problem by J. H. E. Cohn ( "The diophantine equation x^2 + C = y^n", Acta Arithmetica,1993 ). More precisely, x^2+1=y^n has no solution for x, y being positive integers and n greater than or equal to 3. Now, this question is closed. $\endgroup$
    – smart cat
    Commented Feb 16, 2017 at 21:55
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    $\begingroup$ You might add that as an answer. Further, if there are other MathOverflow questions which use the same reference to Cohn for similar questions, you might link to them. Gerhard "Building Better Bridges For Everyone" Paseman, 2017.02.16. $\endgroup$ Commented Feb 16, 2017 at 22:07

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Catalan's Conjecture is a theorem (as of 2002):

With one exception, there are no solutions in positive integers of $$x^a+1=y^b$$ with $a,b \ge 2.$

The exception is, of course, $$2^3+1=3^2$$

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There has been a comprehensive treatment on the solutions of Diophantine equation with the form $x^2+C=y^n$, where $x$ and $y$ are positive integers, $n \ge 3$ and $C$ is a positive integer. (J. H. E. Cohn, "The diophantine equation $x^2+C=y^n$", Acta Arithmetica,1993).

As a special case, when $C=1$, the equation $x^2+1=y^n$ with $n \ge 3$, has no solution.

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