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Topic: Pythagorean factorials (Read 2097 times) |
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JocK
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Pythagorean factorials
« on: May 20th, 2006, 3:54am » |
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0!, 1!, 2! and 6! can all be written as the sum of two perfect squares. How many such 'Pythagorean factorials' exist?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
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Re: Pythagorean factorials
« Reply #1 on: May 20th, 2006, 8:42am » |
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Any number is expressible as the sum of two squares iff its prime factorization does not contain any entries of the form pe where p = 3 mod 4 and e is odd. This makes it very hard for n! to be the sum of two squares, particularly for larger values of n. The 4k+3 primes up to 71 alone show that no other numbers less than 1000 have factorials that are the sum of perfect squares. I suspect that the answer to the question is "4".
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« Last Edit: May 20th, 2006, 9:10am by Icarus » |
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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Eigenray
wu::riddles Moderator Uberpuzzler
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Re: Pythagorean factorials
« Reply #2 on: May 25th, 2006, 12:31pm » |
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There are only finitely many Pythagorean factorials. Since pi3(x) / (x/log x) -> 1/2, where pi3(x) is the number of primes <x which are 3 mod 4, we have [ pi3(x) - pi3(x/2) ] / (x/log x) -> 1/4. Hence for sufficiently large n, there is always a prime p=3 mod 4 with n/2 < p <n, which therefore divides n! exactly once. A constructive bound could be obtained from a careful proof of Dirichlet's theorem, but I'm pretty sure "sufficiently large" here is n>6.
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« Last Edit: May 25th, 2006, 12:37pm by Eigenray » |
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x2862
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Re: Pythagorean factorials
« Reply #3 on: Feb 12th, 2011, 10:25pm » |
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I have listed all the fasctorial from 1 to 2000 on my website http://www.x2862.com/Factorial/ This may or may not be of some use as these are very large numbers and hard to work with.
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