From Pithagoras:

x^2 + (x+2)^2=y^2 where x and y anre ints.

You can investigate only-int solutions. May be a little complicated be should not be impossible.

OR (assuming the +2 side can be any side (hypo.) too)

x^2 + y^2 = (x+2)^2 gives

y^2 = 4X + 4

y = 2 * sqrt(x+1)

x= 8 gives y=6. All are integers.

x= 15 gives y=8, and x+2=17 all integers and fits the answer.

So, there should be infinitly many solution in the above format.

Please note Solution to the first expression may be hard but I suspect if there are infinitely many solutions from: x^2 + (x+2)^2=y^2 probably only 3, 4, and 5. But you got to check that out.

-O