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Topic: Diophantine squares (Read 631 times) |
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NickH
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Diophantine squares
« on: Apr 2nd, 2003, 1:50pm » |
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Show that c2 + 1 = (a2 - 1)(b2 - 1) has no solution in positive integers.
(Edited for towr's (and everyone else's) benefit!)
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| « Last Edit: Apr 3rd, 2003, 10:42am by NickH » |
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towr
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Re: Diophantine squares
« Reply #1 on: Apr 2nd, 2003, 10:15pm » |
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a = b = c = 0 => 1=1
now someone else can try with a,b,c > 0 
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| « Last Edit: Apr 2nd, 2003, 10:15pm by towr » |
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wowbagger
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Re: Diophantine squares
« Reply #2 on: Apr 3rd, 2003, 2:55am » |
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on Apr 2nd, 2003, 10:15pm, towr wrote:| now someone else can try with a,b,c > 0 |
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Let's suppose someone has already shown the nonexistence of solutions with a, b, c > 0. Based on this, I can rule out any solution with any one of a, b, c less than zero.
Does this mean we've already done more than half of the work? 
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LZJ
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Re: Diophantine squares
« Reply #3 on: Apr 3rd, 2003, 5:13am » |
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But that's cheating!!!  It's definitely not half the work, its no work at all.
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SWF
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Re: Diophantine squares
« Reply #4 on: Apr 16th, 2003, 7:26pm » |
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c2+1=(a2-1)(b2-1)=(ab)2- a2-b2+1
c2=(ab)2-a2-b2
As long as a, b, and c are all even divide the equality above by 4 to give new values of a, b, and c satisfying the equality. Repeat until one or more of them is odd.
Odd integers squared have the form 4*n+1, while even integers squared have the form 4*n. With at least one of a, b, or c odd, it can be seen that the two sides of the equality above have a different remainder when divided by 4, so it is not possible for them to be equal. The c2 has remainder 1 or 0. The other term has remainder 3 or 0, and only one side at a time can have remainder 0 since all the common factors of 4 have been removed.
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