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 2021-06-04, 13:04 #1 Charles Kusniec     Aug 2020 Brasil 2·3 Posts 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.
2021-06-04, 14:08   #2
Dr Sardonicus

Feb 2017
Nowhere

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Quote:
 Originally Posted by Charles Kusniec 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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