The discussion focuses on identifying prime numbers p for which the expression p! + p results in a perfect square. The participants confirm that for p = 2 and p = 3, the expressions yield perfect squares. Utilizing Wilson's theorem, they establish that (p-1)! + 1 is divisible by p, leading to the conclusion that k must be a perfect square for the equation p! + p = kp^2 to hold true. As primes increase beyond 3, the factorial grows too large, making it impossible for p! + p to equal p^2.
PREREQUISITESMathematicians, number theorists, and students interested in prime number properties and factorial expressions.
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...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.