 mersenneforum.org > Math Divisibility between polynomial expressions
 Register FAQ Search Today's Posts Mark Forums Read 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

2×2,503 Posts 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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post jnml Miscellaneous Math 6 2020-05-02 12:41 enzocreti FactorDB 2 2018-03-01 21:51 JM Montolio A Miscellaneous Math 3 2018-02-27 16:11 carpetpool Information & Answers 0 2017-11-28 07:44 CRGreathouse Math 3 2009-12-09 19:56

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