MHB Possible webpage title: Which Primes Make p!+p a Perfect Square?

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The discussion focuses on identifying prime numbers p for which the expression p! + p is a perfect square. The initial findings confirm that for p = 2 and p = 3, the expression yields perfect squares. Using Wilson's theorem, the relationship p! + p = p[(p-1)! + 1] is established, indicating that (p-1)! + 1 must be divisible by p. Further analysis suggests that for larger primes, the factorial grows too rapidly, making it unlikely for p! + p to equal a perfect square. The conversation highlights the complexity of the problem and the difficulty in extending the findings beyond small primes.
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Find all prime numbers p for which p!+p is a perfect square.

My thoughts: 2!+2 and 3!+3 are perfect squares.
p!+p=p[(p-1)!+1]
By Wilson's theorem, (p-1)!+1 is divisible by p. Now I'm stuck.
 
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Alexmahone said:
Find all prime numbers p for which p!+p is a perfect square. My thoughts: 2!+2 and 3!+3 are perfect squares. p!+p=p[(p-1)!+1] By Wilson's theorem, (p-1)!+1 is divisible by p. Now I'm stuck.
Well, you are almost done. (p - 1)! + 1 = kp p! + p = kp^2 so we see that k must be a perfect square to satisfy your condition. let k = m^2 p!+p = (mp)^2 also p!+p = c^2 c^2 = (mp)^2 p = c/m now if m = 1 then c = p and p!+p = p^2 that's only going to happen for 2!, 3! cause they have few products. As the primes get larger than 3 the p! has too many products destroying any possibility that p!+p = p^2 Now I'm stuck cause i can't think of a good reason why m can't be any other positive integer...

---------- Post added at 07:05 PM ---------- Previous post was at 07:03 PM ----------

why are all the sentences getting bunched up??
 

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