Question

if p is a prime, prove that p! is not a perfect square. Please, help

Answer 1

My attempt:
let \(p!=\prod_i q_i^{b_i}; i\in \mathbb N \cup\{0\}\)
Suppose \(p! =A^2\)
and \(A=\prod_iq_i^{a_i}\rightarrow A^2=\prod_i q_i^{2a_i} \)
Then for any \(q_i\) on p! , \(\exists q_i \in A^2 s.t ~~q_i^{b_i}=q_i^{2a_i}\)
But p! = p(p-1)(p-2)........2*1
and p is unique, that is exponent of p =1
Then, there is no \(a_i\) s.t \(2a_i =1\)
That shows p! is not a perfect square.
Am I right?

Answer 2

@ganeshie8 @freckles

Answer 3

Looks good to me !
In short, your proof is saying that "\(p!\) is not a perfect square because \(p^2 \nmid p!\)"