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 Sum the squares, get factorial (Posted on 2007-04-03)
Analytically determine all possible triplets (p, q, r) of positive integers that satisfy q≤r and p≤13 and p!= q²+r².

 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.

 Subject Author Date Some insights Jer 2007-04-04 12:10:54 Non-analytical solution Dej Mar 2007-04-03 16:28:51

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