2008-12-18, 06:06 | #1 |
May 2004
100111100_{2} Posts |
Mally's marginal notes
In a book on Einstein - co-authored by many well-known physicists, Mally had noted that 137 ( reciprocal of the constant of fine structure) has a
peculiar property: when multiplied by a natural number, say 8, we get 1096; 10^2 + 96^2 = 9316=68*137. Many products of natural numbers and 137 have similar properties. a) Is there a simple algebraic explanation for this? b) Are there other numbers with similar properties? A.K. Devaraj |
2008-12-18, 07:37 | #2 |
"William"
May 2003
New Haven
3·787 Posts |
It works because 100^{2} is -1 (mod 137).
By construction 100a + b = 0 (mod 137) Hence 100a = -b (mod 137) squaring 10000a^{2} = b^{2} (mod 137) -a^{2} = b^{2} (mod 137) 0 = a^{2} + b^{2} (mod 137) The same kind of trick will work for any factor of 10^{2n}+1. So exactly the same trick will work for 73. ie 73*14=1022; 10^{2}+22^{2}= 584 = 8*73 9901 divides 1000^{2}+1. so 9901*8 = 79208; 79^{2}+208^{2}=49505 = 5*9901 |
2008-12-18, 16:39 | #3 |
"William"
May 2003
New Haven
3·787 Posts |
The trick extends to other powers, too. For x^{3} you need factors of 10^{3n}-1. 37 is the largest factor for n=2, so we get
29*37 = 1073; 10^{3}+73^{3} = 390017 = 10441*37 11*37=407; 4^{3}+7^{3} = 407 = 11*37 42*37=1554; 15^{3}+54^{3}=160839 = 4347*37 37 is also a factor for n=3 (and higher). So the same 37 also gives 29*37 = 1073; 1^{3}+73^{3} = 389018 = 10514*37 42*37=1554; 1^{3}+554^{3}=170031465 = 4595445*37 |
2008-12-19, 03:33 | #4 |
May 2004
2^{2}×79 Posts |
Mally's marginal notes
Tks; I just wanted members to remember him on 18th.
Devaraj |
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