I'll take a stab at it.
Premise: x, y are integers
setup case 1: setup case 2: Continued from
So it can be seen that regardless of which setup case we choose, positive or negative, we end up evaluating the same equation of
Note that if x = 0 then 3 = 0, which is false, so x cannot be zero.
Now, because x and y are both integers, x + 7y must be an integer. Which means that 3/x must be an integer.
So
proof case 1: x = -3
But this contradicts that y is an integer, so
proof case 2: x = -1
But this contradicts that y is an integer, so
proof case 3: x = 1
But this contradicts that y is an integer, so
proof case 4: x = 3
But this contradicts that y is an integer, so
Therefore there are no integer values of x for which y can be an integer also.
So the solution to the problem "find all integer solutions (x, y)" is the empty set.