ben1996123
Banned
Find the number of distinct pairs of integers (x,y) such that 0<x<y and sqrt 1984 = sqrt x + sqrt y.
I tried subtracting sqrt y from each side, then squared the equation, but that just gave me 1984-2sqrt(1984y) + y = x, which I can't really work with.
x = 31, 124, 279
y = 1519, 1116, 775
y = 1519, 1116, 775
sqrt 1984 = sqrt x + sqrt y
sqrt y = sqrt 1984 - sqrt x
y = x - 16sqrt(31x) + 1984
y>x, so y=x+k where k is a constant and k>0
31x is a perfect square
x+k = x-16sqrt(31x)+1984
k = 1984-16sqrt(31x)
1984-16sqrt(31x)>0
124>sqrt(31x)
x < 496
31x is a perfect square and 0<x<496
x = 31, 124, 279
y = 1519, 1116, 775
3 pairs
sqrt y = sqrt 1984 - sqrt x
y = x - 16sqrt(31x) + 1984
y>x, so y=x+k where k is a constant and k>0
31x is a perfect square
x+k = x-16sqrt(31x)+1984
k = 1984-16sqrt(31x)
1984-16sqrt(31x)>0
124>sqrt(31x)
x < 496
31x is a perfect square and 0<x<496
x = 31, 124, 279
y = 1519, 1116, 775
3 pairs