Find all pairs of positive integers, x, y, such that x2 + 3y and y2 + 3x are both perfect squares.
Since x and y are positive, we may write
x2 + 3y = (x + a)2, and
y2 + 3x = (y + b)2
where a, b are positive integers.
Expanding, we find that the squared terms cancel, leaving the linear simultaneous equations
3y = 2ax + a2
3x = 2by + b2
Solving, we obtain
x = (2a2b + 3b2)/(9 − 4ab)
y = (2b2a + 3a2)/(9 − 4ab)
Since a and b are positive, the numerator in each fraction will be positive. For the denominator to be positive, we must have ab = 1 or 2.
If (a,b) = (1,1), (1,2), (2,1), then, respectively, (x,y) = (1,1), (16,11), (11,16). Hence these are the only solutions.
If there is no restriction on the sign of x and y, are there any additional solutions?
Source: Mathematical Diamonds, by Ross Honsberger. See Nine miscellaneous problems.