All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Making it cubic (Posted on 2020-11-10)
Find all the integers n for which (8n-25)/(n+5) is cube of a rational number.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution Comment 1 of 1
Multiply both numerator and denominator by (n+5)^2 to make (8n^3+55n^2-50n-625)/(n+5)^3.  Then all that is needed is for the numerator to be an integer.

The compound inequality (2n+4)^4 < 8n^3+55n^2-50n-625 < (2n+5)^4 is true for most n.  It fails when either (2n+4)^4 >8n^3+55n^2-50n-625 or 8n^3+55n^2-50n-625 > (2n+5)^4.

The first failure inequality is satisfied for n in (-3.97, 24.82) and the second failure inequality is satisfied for n in (-35.81, -4.19).
Then the possible integer n are in (-35, 24).

A quick program finds 8n^3+55n^2-50n-625 is an integer for only n=-5 or n=3.  n=-5 must be discarded as that makes the denominator of the fraction zero.

Thus the only solution occurs when n=3 which yields (8n-25)/(n+5) = (8*3-25)/(8+5) = (-1/2)^3.

 Posted by Brian Smith on 2020-11-21 11:52:47

 Search: Search body:
Forums (0)