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Topic: Improper Fraction Is An Integer (Read 2853 times) |
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ThudnBlunder
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Improper Fraction Is An Integer
« on: Jul 22nd, 2010, 5:47pm » |
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For what positive integer values of n is the improper fraction g(n) = (12n3 - 5n2 - 251n + 389)/(6n2 - 37n + 45) also an integer?
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towr
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Re: Improper Fraction Is An Integer
« Reply #1 on: Jul 23rd, 2010, 1:45am » |
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g(n) = 2n + 1007/(34(2n-9)) - 37/(17 ( 3n-5)) + 23/2 = 0 (mod 1)
1007/(34(2n-9)) - 37/(17 ( 3n-5)) = 1/2 (mod 1)
1007/(17(2n-9)) - 74/(17 ( 3n-5)) = 0 (mod 1)
1007/(2n-9) - 74/( 3n-5) = 0 (mod 17)
If both 1007/(2n-9) and 74/(3n-5) are integer, then n=14 is the only solution.
But while I think it's true that they can't be non-integer, I've yet to find a nice proof of it.
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| « Last Edit: Jul 23rd, 2010, 1:49am by towr » |
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ThudnBlunder
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Re: Improper Fraction Is An Integer
« Reply #2 on: Jul 27th, 2010, 5:39am » |
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g(n) can be rewritten as 2n + 9 + [(3n - 4)(5n + 4)/(3n - 5)(2n - 9)]
(3n - 4) and (3n - 5) being consecutive integers, they are relatively prime
and hence for g(n) be be an integer 3n - 5 must divide 5n + 4.
And when 3n - 5|5n + 4 it also divides 3(5n + 4) - 5(3n - 5) = 37
So 3n - 5 must divide at least one of the divisors of 37, which are 1, -1, 37, and -37.
This gives 2, 4/3, 14, and -32/3 as respective values of n, leaving 2 and 14 as possible answers.
g(2) = 13 - (2*14)/(1)*(-5) = 13 - 28/5 = 37/5 (not an integer)
and
g(14) = 37 + 4 = 41
So g(14) is the only solution, as towr suggested.
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