At the outset, we let x=y= p(say).
Then, we must have:
(p+1)/p = 1 + 1/p is a positive integer, which is possible only when p=1.
Therefore, x=y=1 is a solution.
Now, (x+1)/y is a positive integer
=> x+1 ≥ y => x ≥ y-1 ....(i)
Again, (y+1)/x is a positive integer
=> y+1 ≥ x .....(ii)
Then from, (i) and (ii), we obtain:
y-1 ≤ x ≤ y+1
Since each of x and y is a positive integer, it follows that:
x= y-1, y, y+1
The case x=y has already been considered before.
If x=y-1, then (x+1)/y= 1 and, (y+1)/x is a positive integer means:
(y+1)/(y-1) is a positive integer, so that: 2/(y-1) is a positive integer.
This is possible, when: y=2, giving: x=1, or:
y=3, giving: x=2
If x=y+1 or, y=x-1, then (y+1)/x = 1 and, we consider the restriction (x+1)/y is a positive integer.
Interchanging x and y, we then can refer to the previous case, and obtain two more solutions as:
x=2, y=1 and, x=3, y=2
Consequently, (x,y) = (1,1), (1,2), (2,3), (2,1) and (3,2) gives all possible solutions to the given problem.
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