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

 Mean Muse (Posted on 2014-04-18)
Let gm(x,y) and hm(x,y) respectively denote the geometric mean and harmonic mean of x and y.

Given that each of x, y, gm(x,y) and hm(x,y) is a positive integer:

Can gm(x,y) = hm(x,y)+1?

If so, give an example. If not, prove it.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution Comment 1 of 1
Let gm(x,y) = G, hm(x,y) = H, (x+y) = S.

By definition, G = sqrt(xy) and H = (2*xy)/(x+y).  Then G = H + 1 = (2G^2)/S + 1 giving the quadratic 2G^2 -SG + S = 0 with discriminant D = S^2 - 8S.

Since we're dealing with integers, D is a perfect square so S^2 -8S = k^2, (S-4)^2 = k^2 + 16, and (S-4+k)(S-4-k) = 16.   Possible factors are (16,1), (8,2), (4,4).  The first makes S non-integral, the second gives S=9, the third gives S=8.

For S=9 no choice of (x,y) makes G integral.  For S=8, (x,y)=(4,4) works but then G=H.  So no (x,y) satisfies the problem conditions.

Edited on April 21, 2014, 12:42 pm
 Posted by xdog on 2014-04-21 12:41:00

 Search: Search body:
Forums (1)