$x!+1=y^2$, I found 3 solutions. They are $(4,5),(5,11),(7,71)$. Is there a $4$th solution?If not can you prove it?
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Never mind, not a duplicate, since the OP here asks for just one more solution instead of infinitely many. – Feb 16 '16 at 08:23
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. – JKnecht Feb 16 '16 at 08:33
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I have given a proof in : http://math.stackexchange.com/a/1892315/179940 – Michael Aug 15 '16 at 03:40
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The same question has been asked before here. What you have asked is still an open question in Mathematics.
As stated in the wikipedia article Brocard's Problem, Overholt (1993) showed that there are only finitely many solutions provided that the abc conjecture is true. There is now under evaluation a proof of the abc conjecture by Shinichi Mochizuki.
Anirban Mandal
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