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Topic: Sum Squared Integers And Divide, Get Power (Read 537 times) |
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K Sengupta
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Sum Squared Integers And Divide, Get Power
« on: Jul 27th, 2007, 8:12am » |
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Analytically determine all possible quadruplets of positive integers (a, b, c, d) with a<=b<=c<=d satisfying:
a2 + b2 + c2 + d2 = 7*4p, where p is a non-negative integer.
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| « Last Edit: Jul 27th, 2007, 9:20am by K Sengupta » |
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pex
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Re: Sum Squared Integers And Divide, Get Power
« Reply #1 on: Jul 27th, 2007, 9:02am » |
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on Jul 27th, 2007, 8:12am, K Sengupta wrote:Analytically determine all possible quadruplets of non-negative integers (a, b, c, d) satisfying:
a2 + b2 + c2 + d2 = 7*4p, where p is a non-negative integer.
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Aren't there infinitely many solutions, at least one for each p? As far as I recall, every positive integer is the sum of four squares (counting 0 = 02 as a square, which is not excluded in your problem statement).
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K Sengupta
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Re: Sum Squared Integers And Divide, Get Power
« Reply #2 on: Jul 27th, 2007, 9:11am » |
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on Jul 27th, 2007, 9:02am, pex wrote:
Aren't there infinitely many solutions, at least one for each p? As far as I recall, every positive integer is the sum of four squares (counting 0 = 02 as a square, which is not excluded in your problem statement). |
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True.
The quadrupets (a, b, c, d) of integers should be positive instead of non- negative in Para 1 of the problem statement.
I confirm having corrected the said anomaly.
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| « Last Edit: Jul 27th, 2007, 9:14am by K Sengupta » |
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Eigenray
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Re: Sum Squared Integers And Divide, Get Power
« Reply #3 on: Aug 5th, 2007, 9:54am » |
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p=0: (1,1,1,2)
p=1: (2,2,2,4), (1,1,1,5), (1,3,3,3)
If p>1, then a,b,c,d must all be even, since any square is congruent to 0,1, or 4 mod 8. Dividing the equation 22 therefore gives a bijection between solutions for p and p-1. So the only solutions are 2p(1,1,1,2), 2p-1(1,1,1,5), and 2p-1(1,3,3,3)
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