# Number Theoryp!+p is a perfect square

#### Alexmahone

##### Active member
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.

Last edited:

#### agentmulder

##### Active member
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 thats 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??