Sam Loyd's Missing Square

Paradox
Consider a square of side length $13$.

Let it be divided into:
 * a $13 \times 5$ rectangle, divided into two right triangles by its diagonal
 * a $13 \times 8$ rectangle, divided into trapezia with side lengths $8$ and $5$:


 * MissingSquare1.png

Let the pieces be rearranged to form two long right triangles arranged to form a $21 \times 8$ rectangle.

The area of the square is $13 \times 13 = 169$.

The area of the rectangle is $21 \times 8 = 168$.

Where did the missing $1 \times 1$ square go?

Resolution
This is a falsidical paradox.

When you place the $13 \times 5$ right triangle against the trapezium, supposedly to make a $21 \times 8$ right triangle, the hypotenuse of that figure is not actually straight.

When the $21 \times 8$ rectangle is drawn accurately, you will see an overlap:


 * MissingSquare2.png

It is noted that $5$, $8$ and $13$ are consecutive Fibonacci numbers.

From Cassini's Identity:


 * $F_n^2 = F_{n - 1} F_{n + 1} \pm 1$

Hence the resolution.