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Does anyone know how to prove there is never a perfect square for N! when N  > 2

 Sep 28, 2015
 #1
avatar+128053 
+5

Here's my attempt at this one......

 

Let's suppose that N!   = p^2

 

And  every square can be factored into a product of squared primes

 

Then....let's suppose that  p is factored as  (a1)2 * (a2)2 * (a3)2 * .....(an)2   where each term is a squared prime

 

And N! = 2 * 3 * 4 *......*N-2 * N-1 * N

 

So we have that........

 

2 * 3 * 4 *......*N-2 * N-1 * N  = (a1)2 * (a2)2 * (a3)2 * .....(an)2    

 

But some of the terms on the left side will not be perfect squares, while every term on the right hand side will be a perfect square

 

Thus.......the left side cannot be written solely as a product of perfect square primes

 

Thus.......N!  is not a perfect square

 

 

cool cool cool

 Sep 29, 2015
edited by CPhill  Sep 29, 2015
edited by CPhill  Sep 29, 2015
 #2
avatar+89 
0

Thank you CPhill

I posted the question but it did not post with my account credentials.

 Sep 29, 2015

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