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Topic: Divide Two Sets Of Quadratic, Get Integers (Read 259 times) |
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K Sengupta
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Divide Two Sets Of Quadratic, Get Integers
« on: Jul 24th, 2007, 8:01am » |
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Analytically determine all possible pairs of positive integers (a, b) such that :
(a + b2)/(a2-b) and (a2+b)/(b2-a) are both positive integers.
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| « Last Edit: Jul 24th, 2007, 8:17am by K Sengupta » |
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Eigenray
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Re: Divide Two Sets Of Quadratic, Get Intege
« Reply #1 on: Jul 26th, 2007, 2:41pm » |
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| hidden: | If a=b, then (a+a2)/(a2-a) = (a+1)/(a-1) = 1 + 2/(a-1), so a-1 | 2, so we have a=b=2 or a=b=3.
Otherwise, WLOG suppose a<b.
(a2+b)/(b2-a)  1, so if a  (b-2), then we would have
b2-(b-2) b2-a a2+b (b-2)2+b,
or 2b 2, giving b=1 and a<0, a contradiction.
So we must have a=b-1. So (b2+b-1)/(b2-3b+1) is an integer. Since it's clearly > 1, it must be at least 2, so
b2+b-1 2(b2-3b+1), or
b2-7b+3 0,
which gives b < 7. Checking b=3,4,5,6, the only solution is (a,b) = (2,3). |
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