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l4z3r
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 Integral Solutions   « on: Aug 29th, 2008, 4:46am » Quote Modify

a function is defined as:

f: Z(+)  -->  Z

f(m,n) =  (n3 + 1)/ (mn - 1)

where Z(+) denotes the set of positive integers and Z the set of integers.

Find all the solutions for (m,n)

EDIT: f(m,n) not f(x)
 « Last Edit: Aug 29th, 2008, 5:42am by l4z3r » IP Logged
towr
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 Re: Integral Solutions   « Reply #1 on: Aug 29th, 2008, 5:23am » Quote Modify

Shouldn't the x in f(x) come into it somewhere?
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l4z3r
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 Re: Integral Solutions   « Reply #2 on: Aug 29th, 2008, 5:42am » Quote Modify

ah. meant f(m,n). sorry.
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SMQ
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 Re: Integral Solutions   « Reply #3 on: Aug 29th, 2008, 5:58am » Quote Modify

So, in other words, "find all , such that ( + 1) / ( - 1) ", right?

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l4z3r
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 Re: Integral Solutions   « Reply #4 on: Aug 29th, 2008, 7:00am » Quote Modify

yes. hint:use n3+1   1(mod3) and mn-1   -1 (mod n) (number theory)
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Eigenray
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 Re: Integral Solutions   « Reply #5 on: Aug 29th, 2008, 11:53am » Quote Modify

If (m,n) is a solution, then (m, (m2+n)/(mn-1)) is also; then use the ideas that appear here (and which should appear here).

Or, you can proceed more directly by writing n3+1 = (mn-1)((an-m)n-1) and bounding.
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l4z3r
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 Re: Integral Solutions   « Reply #6 on: Aug 30th, 2008, 5:34am » Quote Modify

hmm. good one. I agree with the first part.

If (m,n) is a solution, then (m, (m2+n)/(mn-1)) is also

but, instead of (mn-1)((an-m)n-1) i feel a better alternative would be (kn-1)(mn-1). Gives the answer in lesser steps, i think.
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