To Coda:
According to the Pythagorean theorem, the sum of the areas of squares on the legs of a right triangle equals the area of the one on its hypotenuse. But the same is true if one builds pentagons or $n$-gons on the sides instead of squares.
There are cases when all three square areas are integers (for Pythagorean triples, for example). Are there such cases for other polygons?
Coda:
NOTE: my original question was the one below
Given the polygonal number formula: $(n^2*(s-2) - n*(s-4))/2$, one can list, for example, the square numbers (s=4): $f(n) = [1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81, ...]$ or the pentagonal numbers (s=5): $f(n) = [1 , 5 , 12 , 22 , 35 , 51 , 70 , 92, ...]$. By inspection we see for the pentagonal numbers that f(4) + f(7) = f(8). Is this analogous to the Pythagorean theorem? Where f(3)+f(4)=f(5).
The area scale factor between the square and the other polygon is the same for all sides. Thus, whatever polygons you put on the sides instead of squares with integer areas, you can scale the entire triangle by the square root of that factor to make all the polygon areas come out to the same integers as the squares did.