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Solve $x! +1 = y^2$ over integers.

I don't know what to try here. The equation implies that $x! = (y-1)(y+1)$ but I'm not sure if that helps at all.

Giorgos Giapitzakis
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Tatarusan Marius
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    $x=4$, $y=5$ works. ($4! + 1 = 25 = 5^2$). Just an example. Also, $x= 5!, y= 11$ Works. $5!+1 = 121= 11^2$. – amWhy Mar 08 '20 at 16:58
  • Oops, in the second example, I meant $x= 5, y=11$. – amWhy Mar 08 '20 at 17:04
  • The Question as it relates to nonnegative integers (note $x\in \mathbb N$ implies $x!+1$ is positive) is known as Brocard's problem. Any solution $(x,y)$ in that context gives rise to the corresponding solution $(x,-y)$. The only further possibility would involve negative integers $x$. But the factorial function $x!$ does not give an integer result for negative integers $x$, as the gamma function has analytic poles there. – hardmath Mar 08 '20 at 19:50

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The three known solutions $(x,y)$ to Brocard's problem are $(x,y)=(4,5)$, $(x,y)=(5,11)$, and $(x,y)=(7,71)$.

amWhy
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Geoffrey Trang
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  • Sort of a "link only" answer (though I had to provide the link). Please explain (summarize) why no further solutions have been known. – amWhy Mar 08 '20 at 17:05
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    Considering "Paul Erdős conjectured that no other solutions exist. Overholt (1993) showed that there are only finitely many solutions provided that the abc conjecture is true. Berndt & Galway (2000) performed calculations for n up to 109 and found no further solutions. Matson (2017) has recently claimed to have extended this by 3 orders of magnitude to one trillion" I'd say "Please explain (summarize) why no further solutions have been known" is a bit of an unrealistic request. – fleablood Mar 08 '20 at 17:13
  • Thanks , I didn't know this was an unsolved problem. It's in a math magazine with olympiad problems. – Tatarusan Marius Mar 08 '20 at 17:16