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Consider $4!=24$, if you add one you get $25=5^2$. The same occurs with $5! = 120 = 11^2 - 1$, and $7! = 5040 = 71^2 - 1$. Are there other solutions of the equation $n!+1 = m^2$?

I verified that no other solution exists with $m<10^9$. Does the problem has already been studied? Is there a demonstration that no other solution exists?

Greetings.

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It's probably discussed in Guy, Unsolved Problems In Number Theory (I don't have the book at hand, so cannot check). –  Gerry Myerson Nov 6 '10 at 21:43
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You might enjoy this one, $$ $$ mathoverflow.net/questions/16341/… $$ $$ Gerry, it is in Guy, section D25, page 193 in the second edition. About 8 references. $$ $$ –  Will Jagy Nov 6 '10 at 22:29
    
Thank you, i'll check those documents. –  Daniele Morelli Nov 7 '10 at 0:14

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