How many pairs of integers (x,y) satisfy the eqn sqr root x + sqr root y =sqr root 200600?

Answer 1

I was thinking about the square root values need to be the same, but that's not the case since square root of 200600 is not an integer either, I essentially come with the same scheme as kempos, but correct answers for them. In order for the values of the square root to be "exactly" equal, the values must equate analytically.

Therefore, breaking down √200600, you get 10√2006, and √x + √y must equal 10√2006, so...
√x = √2006, √y = 9√2006
√x = 2√2006, √y = 8√2006
√x = 3√2006, √y = 7√2006
√x = 4√2006, √y = 6√2006
√x = 5√2006, √y = 5√2006, for the total of 5 pairs (not counting the repeated ones) Then
(x, y) = (2006, 162486), (8024, 128384), (18054, 98294), (32192, 72216), (50150, 50150)


XR

Answer 2

I got 9, but not sure.

sqrt200600 = 10sqrt2006

let x = sqrt(a^2*b)
y = sqrt(a^2*b)

sqrt(x) + sqrt(y) = sqrt(a^2*b) + sqrt(a^2*b) = a*sqrt(b) + c*sqrt(b) = sqrt(b) * (a + c)
b = 2006, a + c = 10
therefore 1 and 9; 2 and 8, 3 and 7, 4 and 6, 5 and 5.
Here are the pairs:
(2006, 9*2006) = (2006, 18054)
(9*2006, 2006) = (18054, 2006) and so on...
(2*2006, 8*2006)
(8*2006, 2*2006)
(3*2006, 7*2006)
(7*2006, 3*2006)
(4*2006, 6*2006)
(6*2006, 4*2006)
(5*2006, 5*2006)