Four Right Triangles Make a Quadrilateral

In the diagram above, we have that $w^2+y^2=(a^2+b^2)+(c^2+d^2)=(b^2+c^2)+(d^2+a^2)=x^2+z^2$ Thus we are looking for the smallest four distinct integers such that the sum of squares of two of them equals the sum of squares of the other two. This is akin to finding an integer that can be represented as the sum of two perfect squares in two different ways.

A quick perusal of the first few perfect squares shows us that the lowest such integer is $50=1^2+7^2=5^2+5^2$ ; however it doesn't work in this case as we need four distinct integers. The next lowest such integer is $65=1^2+8^2=4^2+7^2$ , which gives us a perimeter of $1+4+7+8=\boxed{20}$

If $a, b, c, d$ are the integer hypotenuses of the quadrilateral in rotation, then this condition must be satisfied:

$a^2 - b^2 + c^2 - d^2 = 0$

or

$- a^2 + b^2 = c^2 - d^2$

This can be worked out by letting one of the 4 interior sides be the unknown $x_1$ , and successively find the rest of the interior sides in terms of $x_1$ until it reaches the original side of length $x_1$ . $x_1$ then drops out and the condition for $a, b, c, d$ then arises.

$x_2 = \sqrt{a^2-x_1^2}$
$x_3 = \sqrt{b^2-x_2^2}$
$x_4 = \sqrt{c^2-x_3^2}$
$x_1 = \sqrt{d^2-x_4^2}$

The smallest distinct set meeting the condition is $1, 4, 8, 7$
To find this, consider

$(-a+b)(a+b)=(c-d)(c+d)$

Then the smallest nonsquare odd composite number factorizable in two ways is $15$

$15 = 3 \cdot 5 = 1 \cdot 15$
$(-1+4)(1+4) = (8-7)(8+7)$

A quadrilateral formed with these sides is not unique.