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Old 2021-06-04, 13:04   #1
Charles Kusniec
 
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Aug 2020
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Default Divisibility between polynomial expressions

Given the expression
\[x = \frac{n^2 -bn}{(2n+k)-b}\]
For \(b\) an integer coefficient, \(k\) is any integer constant, and index −∞<\(n\)<∞.
What is the relationship that must exist between the coefficient \(b\) and constant \(k\), \(x\) will be a sequence of integers generated by the index \(n\)?

I suppose it is better to divide the solutions between \(k\)=even and \(k\)=odd.
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Old 2021-06-04, 14:08   #2
Dr Sardonicus
 
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Quote:
Originally Posted by Charles Kusniec View Post
Given the expression
\[x = \frac{n^2 -bn}{(2n+k)-b}\]
For \(b\) an integer coefficient, \(k\) is any integer constant, and index −∞<\(n\)<∞.
What is the relationship that must exist between the coefficient \(b\) and constant \(k\), \(x\) will be a sequence of integers generated by the index \(n\)?

I suppose it is better to divide the solutions between \(k\)=even and \(k\)=odd.
(n^2 - b*n)/(2*n+k-b) may be rewritten as

n/2 - (b+k)/4 + (k^2 - b^2)/(8*n+4*k-4*b)

The only way to make the expression a fraction whose denominator remains bounded as n increases without bound, is to take k = b or k = -b. If k = b the expression is n/2 + b/2. If k = -b the expression is n/2. Either way, the expression will only be an integer for half the integer values of n.
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