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Topic: Divide The Expressions, Get Integers (Read 276 times) |
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K Sengupta
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Divide The Expressions, Get Integers
« on: Jul 24th, 2007, 8:14am » |
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Analytically determine all possible pairs of positive whole numbers (p, q) such that:
(p2q + p + q)/(pq2 + q + 7) is a positive integer.
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| « Last Edit: Jul 24th, 2007, 8:25am by K Sengupta » |
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Eigenray
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Re: Divide The Expressions, Get Integers
« Reply #1 on: Jul 26th, 2007, 3:18pm » |
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Let k = (p2q+p+q)/(pq2+q+7). Then| hidden: | qk - p = (q2-7p)/(pq2+q+7)
is an integer. Clearly it is < 1.
Case (I): qk-p  -1. That is,
7p-q2  pq2+q+7,
so in particular 7p > pq2, which means q is either 1 or 2.
If q=1, then we have
qk-p = (1-7p)/(p+8) = -7 + 57/(p+8)
is an integer, so p+8 | 57. The only divisors of 57 larger than 8 are 19 and 57, so we get the solutions (p,q) = (11,1) and (49,1).
If q=2, we similarly have 4(qk-p) = -7 + 79/(4p+9) is an integer. But 79 is prime, and 4p+9 can't be 1 or 79.
Case (II): qk-p=0. Then q2=7p, and k=p/q=q/7. This gives the infinite family of solutions (p,q) = (7k2, 7k). |
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| « Last Edit: Jul 26th, 2007, 3:20pm by Eigenray » |
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