Here is one way.Originally Posted by jzon
y can only be 0 or +,-1, because if y were |2| or greater, then 5y^2 is already 20 or more---which is greater than 14.
If y=0, then x^2 = 7, and x is not an integer.
If y= |1|, then x^2 = 9/2, and x is not an integer.
x can only be 0, |1|, or |2|. If x were |3| or more, then 2x^2 is already 18 or more.
If x=0, y^2 = 14/5 and y is not an integer.
If x=|1|, y^2 = 12/5 and y is not an integer.
If x=|2|, y^2 = 16/5 and y is not an integer.
Therefore, there is no solution in x and y as integers.