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.
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, (S4)^2 = k^2 + 16, and (S4+k)(S4k) = 16. Possible factors are (16,1), (8,2), (4,4). The first makes S nonintegral, 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 20140421 12:41:00 