Pythagorean Proof Problem

Let the white triangles have shorter leg $x$ so the longer leg is $x+1$ and hypotenuse $\sqrt{2x^{2}+2x+1}$ .

The green triangles then have shorter leg $2x$ and longer leg $\sqrt{2x^{2}+2x+1}+2x$ .

The Pythagorean Theorem on the green triangle gives the equation $(2x)^{2}+(\sqrt{2x^{2}+2x+1}+2x)^{2}=157$ .

Which has positive solution $x=3$ so the white triangle is a $3-4-5$ and the area sought is $4\cdot\frac{1}{2}\cdot3\cdot4=\boxed{24}$

solve 4x^2+(y+2x)^2=157, y^2=x^2+(1+x)^2, z=y^2-1, x=3, y=5 and z=24

Answer=24

Let us call the sides of the green triangles, A, B and C, and those of the white triangles, a, b and c

Thus from Pythagoras we have

A^2 + B^2 = C^2 where C^2=157

and

a^2 + b^2 = c^2

Let us assume an integer solution to the problem, as if c^2 and therefore c are irrational numbers, it is unlikely that C^2 would be an integer.

Based on this assumption, let us see if 157 can be written as the sum of two integer squares. It can - A^2 = 121 and B^2 = 36.

Therefore A = 11 and B = 6.

c = A - B = 11-6 = 5

c^2 = 25 and therefore the white area = 25 - 1^2 = 24.

(As a check, c^2 = 25 = 9 + 16. Therefore a and b are 3 and 4 respectively, which is consistent with the side of the red square being 1).