perplexus.info :: Numbers : Three questions plus one challenge

Three questions plus one challenge (Posted on 2018-01-04)

Solve the equation:
n! + 1 = m²

given n and m are integers below 100

AFAIK there are 3 solutions in this range.

Bonus: Explore bigger numbers and post your comments.

No Solution Yet Submitted by Ady TZIDON

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Probably done

Comment 1 of 1

4! + 1 = 25 = 5^2

5! + 1 = 121 = 11^2

7! + 1 = 5041 = 71^2

https://oeis.org/A146968 gives the short sequence and notes there are no other solutions with n<10^9 (A fair bit higher than 100)

It also has some links.

A solution occurs whenever http://oeis.org/A038202 equals 1. The speed at which this sequence grows is a strong hint that there are no other solutions.

Posted by Jer on 2018-01-04 09:59:54

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