TITLE: The number $\pi$ and summation by $SL(2,\mathbb Z)$
QUESTION [114 upvotes]: Let $f(a,b,c,d)=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}-\sqrt{(a+c)^2+(b+d)^2}$. (it is the defect in the triangle inequality) 
Then, we discovered by heuristic arguments and then verified by computer that
$$\sum f(a,b,c,d)^n = 2-\pi/2$$
where the sum runs over all $a,b,c,d\in\mathbb Z$ such that $a\geq 1,b,c\geq 0, ad-bc=1$ and $n=2$.
It seems that when $n=1$, we obtain $2$ in the right hand side. We have failed to guess the result for $n>2$.
So the question is: can you prove the result for $n=2$? (We can, but in a rather unnatural way. We will write this later and now we want to tease the community, probably somebody can find a beautiful proof.) 
Added: we have two answers for the above question, so the rest is:
Question: Can you guess the result for $n>2$? I tried http://mrob.com/pub/ries/ but nothing interesting was revealed.
PS. In case it can help someone, below are these sums of powers ($n=1,2,3,4,5$), calculated by computer:
$1.9955289122768913 = 2$
$0.4292036731309361 = 2 - \pi/2$
$0.21349025954227965 = $ ?
$0.11983665032283052 = $ ?
$0.06933955916793563 = $ ?
Added: this and something more can be found in https://arxiv.org/abs/1701.07584 and https://arxiv.org/abs/1711.02089

REPLY [26 votes]: I add our explanation and the origin of the problem. 
To obtain the formula we just need to verify the following lemma (by a   straightforward computation). 
Magic Lemma. Let $(a,b),(c,d)$ be as in our summation in the main posting. Draw tangents to the unit circle which are orthogonal to the directions $(a,b),(c,d),(a+c,b+d)$. Then the triangle in the intersection has the area $\frac{f(a,b,c,d)^2}{2}$. 
Now consider the circle given by $x^2+y^2=1$. At every rational point on it we draw the tangent line. Clearly, our disk is the intersection of all the half-planes given by these lines.
So we can approximate the area of the disk by the following sequential cutting. We start with the square $[-1,1]\times [-1,1]$. It has four points of tangency with our circle. We will construct a sequence of polygons converging to the circle. Initially we have sides of directions orthogonal to $(1,0),(0,1),(-1,0),(0,1)$. At every step, we take two vectors $v_1,v_2$(in the same quadrant) which give the basis of $\mathbb Z^2$ and add to our polygon a new side, which is given by the tangent line of the direction orthogonal to $v_1+v_2$.
For example, the first step will be cutting the square by the line $x+y=\sqrt 2$, if we take the vectors $(1,0),(0,1)$. The area of the eliminated triangle is $\frac{(\sqrt2-1)^2}{2}=\frac{f(1,0,0,1)^2}{2}$.
It seems that this procedure can be generalized to higher dimensions — at least a preliminary computation shows that something like the above magic lemma should happen.