Expert: Paul Klarreich - 12/2/2009

Question
y^2=x! + 24   x,y are natural numbers
y^2 = x! +4!
y^2=(x(x-1)(x-2)...(x-(x-1))+4*3*2*1
y = sqrt( (x(x-1)(x-2)...(x-(x-1))+24)
ok now I am stuck
btw I know that (5,12) satisfy this because I tried all the numbers from 1 till I stopped at 5 but I need an algebraic solution  

Answer
Questioner: hamad
Country: Kuwait
Category: Advanced Math
Private: No
Subject: algebra factorial problem
Question: y^2=x! + 24   x,y are natural numbers
y^2 = x! +4!
y^2=(x(x-1)(x-2)...(x-(x-1))+4*3*2*1
y = sqrt( (x(x-1)(x-2)...(x-(x-1))+24)
ok now I am stuck
btw I know that (5,12) satisfy this because I tried all the numbers from 1 till I stopped at 5 but I need an algebraic solution
........................
There might not be one.


y^2  = (x(x-1)(x-2)...3 2 1) + 4 3 2 1   << your equation
  
=  [x(x-1)(x-2)...5 + 1] 4 3 2 1   << factor out the 4 3 2 1

Now the bracketed stuff must have 3 2 1 in it, because 4 is a square.

So x = 5 works because  5 + 1 = 6 = 3*2.

But you could probably prove that no other number works.