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Factorial and a Square (Posted on 2008-02-23) Difficulty: 2 of 5
If a and p are positive integers, then solve for (a,p):
a2=p!

See The Solution Submitted by Praneeth    
Rating: 2.0000 (4 votes)

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re: No Subject | Comment 2 of 9 |
(In reply to No Subject by Dej Mar)

According to the Chebyshev's Theorem that's also known as Bertrand's Postulate, (there's another Chebyshev's Theorem that's not related) there is always a prime number between any number larger than 1 and its double (i.e., between n and 2n for n>=2).

So between the first appearance of a given prime in the formation of n! as 1x2x3x..., and its second appearance as a factor (as 2n), there is always some other prime, by its lonesome, in between, as 2n will come before 2p (where p is that other prime).  So there will always be an unpaired prime in the construction of any factorial, and so the factorial can't be a perfect square so long as it's greater than 1.


  Posted by Charlie on 2008-02-23 14:54:44
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