Tiling of squares in instances of Pythagoras Theorem The Pythagorean Theorem (PT) states that in a right triangle, the square on the hypotenuse equals the sum of the squares on the legs or $a^2 + b^2 = c^2$.
Is there an instance of PT (a given $a$, $b$ and $c$) where there exists a tiling of all three squares in congruent shapes? 
Well, yes, there is one, where $a = b$:

where it is easy to count the smaller triangles in each square by inspection and add $8 + 8 = 16$. The tiles only happen to be similar to the original triangle.
I am curious if there is one where $a < b$, or a proof (or proof idea) that one cannot exist. The tiles can be similar to the triangle or not, as long as they are all congruent. The idea is to give a simple verification of the single PT instance by counting the tiles.
 A: For a given Pythagorean triangle with shorter leg $a$ (e.g., vertical) and longer leg $b$ you could build a grid composed by small right triangles with legs $\frac{1}{b}, \frac{1}{a}$. These small triangles are similar to the initial triangle because the ratio between the smaller and the longer leg is $\frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}$. Each pair of these small triangles forms a small $\frac{1}{b} \cdot \frac{1}{a}$ rectangle, and arranging $ab$ of these rectangles in $b$ rows and $a$ columns (with the rectangles having the smaller side $\frac{1}{b}$ in the vertical direction) we can get a unitary square. In such a grid, each unitary square contains $2ab$ triangular tiles that are similar to the original triangle, and all three squares built on the sides of the initial triangle (including that drawn from the hypothenuse) contain an integer, countable number of tiles. The following image shows an example for the Pythagorean triangle $3,4,5$, where the small triangles have legs $1/4$ and $1/3$, respectively, and where the unitary square contains $3\cdot4=12$ small rectangles and $24$ small triangles.

Another example that shows how to tile the square on the hypotenuse in a way that is easy to confirm is an accurate tiling and is the correct sum.

The triangle is 1, 2, $\sqrt{5}$. The tiles are all of size 1/2, 1, $\sqrt{5}/2$. The smaller square is 1x1, the larger is area 4, and the square on the hypotenuse is area 5 by adding the unit square in the middle, and four of the area 1 triangles around.
This tiling has the basic idea of a proof of the full general PT using the algebraic identity: 
$$
\begin{array}{rcl}
{\rm the\ sum\ of\ the\ areas\ of\ the\ two\ leg\ squares} && a^2 + b^2\\
{\rm the\ center\ full\ unit\ squares}  &=& (a−b)^2\\
{\rm the\ areas\ of\ the\ 4\ triangles}  & & + 4 (a b)/2\\
{\rm the\ area\ of\ the\ square\ on\ the\ hypotenuse} &=& c^2
\end{array}
$$
But the point to the tiling is that the area of the square on the hypotenuse, without such a nice tiling (or a proof of PT) is inscrutable because you can't just 'see' that the areas sum just right (how does your eye cut up the pieces). The tiling here allows you to at least count the tiles, presumably a more trustworthy operation than eye-balling the area.
