All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Sum the squares, get factorial (Posted on 2007-04-03) Difficulty: 3 of 5
Analytically determine all possible triplets (p, q, r) of positive integers that satisfy q≤r and p≤13 and p!= q+r.

  Submitted by K Sengupta    
Rating: 3.0000 (1 votes)
Solution: (Hide)
We now that a positive integer can be expressed as sum of two squares if and only if each of its prime factors of the form 4k+3 occurs as an even power. It is given that p is a positive integer. So, p cannot have a prime divisor of the form 4k+3 raised to an odd power.

So, for p = 3, 4, 5 the highest power of 3 dividing p! is 3^1.

For p = 7, 8, 9, 10, 11, 12, 13 the highest power of 7 dividing p! is 7^1.

So in all those cases n! cannot be written as a sum of two squares.

For p=1, we would obtain r^2 + s^1 =1, which does not possess any solution in positive integers.

For p=2; we obtain 1^2 + 1^2 =2, giving (q,r) = (1,1)

For p =6, we obtain 720 =12^2 + 24^2, giving (q,r) = (12, 24)

Thus, (p,q,r) = (2,1,1);(6,12,24) are the only possible solutions to the problem.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSome insightsJer2007-04-04 12:10:54
Non-analytical solutionDej Mar2007-04-03 16:28:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information