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.
When we say that "P+p is a perfect square," it means that the sum of two numbers, P and p, can be written as the square of another number.
To determine if P+p is a perfect square, you can take the square root of the sum and see if it is a whole number. If it is, then P+p is a perfect square.
If you know that P+p is a perfect square, you can use algebra to solve for the values of P and p. You can set up an equation such as (x+y)^2 = P+p and solve for x and y.
An example of P+p being a perfect square is when P=4 and p=9. The sum of these two numbers is 13, and the square root of 13 is approximately 3.61, which is a whole number. Therefore, 4+9=13 is a perfect square.
P+p being a perfect square has various applications in mathematics and science. It can be used to solve equations, find the length of sides in geometric shapes, and even in cryptography. It is also a fundamental concept in number theory.