(x,y) = (27,11),(11,27),(192,77),(77,192),(4,4),(69,56) and (56,69) constitutes all possible pairs satisfying conditions of the problem.
EXPLANATION:
We know that both x and y are positive. Accordingly, without loss of generality (WLOG), we obtain:
x^2 + 5y = (x + p)^2, and
y^2 + 5x = (y + q)^2
--------------(#)
Now, solving for x^2 + 5y = (x + p)^2; we obtain:
p^2+ 2px - 5y=0 , giving p=-x+/-sqrt(x^2+5y).
Since, x^2+5y > x^2, it follows that exactly one of the roots corresponding to p is positive, while the other root is negative. Since, both the said roots yield the same value for x^2 + 5y, WLOG, we can assume that p is a positive integer.
By similar reasoning, it can be assumed that q is a positive integer.
Simplifying (#), we obtain:
5y = 2px + p^2
5x = 2qy + q^2
Solving, we observe that:
x = (2*(p^2)q+ 5*(q^2))/(25 – 4pq)
y = (2(q^2)p + 5*(p^2))/(25 – 4pq)
Since p and q are positive, the numerator in each of the abovementioned fractions must be positive. For the denominator to be positive, we must have pq = 1,2,3,4,5,6.
If (p,q) = (1,1), (1,2), (2,1),(1,3),(3,1), (1,4),(4,1)`(1,5),(5,1),(1,6),(6,1), (2,2),(2,3), (3,2) then respectively:
(x,y) = (1/3, 1/3),(24/17, 13/17),(13/17, 24/17), (51/13, 23/13), (23/13, 51/13), (88/9, 37/9), (37/9, 88/9), (27, 11),(11,27), (192, 77),(77,192),(4,4), (69, 56),(56,69).
It is clearly observed from above that (27,11),(11,27),(192,77),(77,192),(4,4),(69,56) and (56,69) constitutes all possible positive integral pairs corresponding to (x, y).
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