# Part 2: Does the arithmetic mean of sides right triangles to the mean of their hypotenuse converge?

A primitive Pythagorean triplet is a triplet $$a^2 + b^2 = c^2$$ be where $$a,b,c$$ have no common factors and is generated by $$a = r^2 - s^2, b = 2rs, c = r^2 + s^2$$ where $$r > s \ge 1, \gcd(r,s) = 1$$ and exactly one of the two numbers $$r$$ and $$s$$ is even. Clearly as $$r$$ increases, the number of primitive triplets formed for a given $$r$$ increases since the number of $$s$$ satisfying the above conditions increases.

Claim: Let $$c_1,c_2,\ldots$$ be the hypotenuse and $$b_1,b_2,\ldots$$ be the corresponding longer of the two orthogonal sides formed for Pythagorean triangles for all $$r \le x$$ then as $$x \to \infty$$,

$$\frac{b_1 + b_2 + b_3 + \cdots}{c_1 + c_2 + c_3 + \cdots} = \sqrt{2} - \frac{1}{2}$$

Can this claim be proved or disproved?

The difference between this question and the related question: Part 1: Does the arithmetic mean of sides right triangles to the mean of their hypotenuse converge? is that here the triangles are in sequenced in ascending order of $$r$$ and $$s$$ where as in the related question, they are sequenced in ascending order of the hypotenuse and depending on the choice of sequencing, the limiting value differs.

• @Aretino Here is a similar problem with a different sequencing order of the triplets – Nilotpal Kanti Sinha Oct 25 '19 at 10:27
• You forgot to add that $r$ and $s$ cannot both be odd. Moreover, it is not clear (in my opinion) how you order the triples. – Aretino Oct 25 '19 at 16:08
• @Aretino Added that now. Regarding order, we just need to generate all triplets for $r \le x$ for some $x$ and make $x \to \infty$ – Nilotpal Kanti Sinha Oct 25 '19 at 16:16
• You should add that all sides are integer numbers. – Ripi2 Oct 25 '19 at 16:55
• @Ripi2 That is already in the word Pythagorean triplet because by definition all numbers in a Pythagorean triplet are natural numbers – Nilotpal Kanti Sinha Oct 25 '19 at 22:57

Your claim is right. To see why, consider the quantities $$B_r=\sum_{\text{constant }r} b_{r,s}, \quad C_r=\sum_{\text{constant }r} c_{r,s}.$$ I will show that $$B_r/C_r$$ tends to $$\sqrt2-1/2$$ for $$r\to\infty$$.
Given any integer $$r\ge2$$ the possible values for $$s$$ are $$r-1$$, $$r-3$$, $$r-5$$, ... as long as $$s>0$$. We can summarise that as follows: $$s=r-2k+1,\quad\text{where}\quad 1\le k\le \lfloor{r/2}\rfloor.$$ We must take into account that $$b_{r,s}$$ can be given by two different expressions: $$b_{r,s}=\max(r^2-s^2, 2rs)= \cases{ 2r(r-2k+1) & for 1\le k where $$N(r)={2-\sqrt2\over2}r+{1\over2}$$ is the value of $$k$$ for which $$r^2-s^2=2rs$$. We may then compute $$B_k$$ as follows: $$B_k=\sum_{k=1}^{\lfloor{N(r)}\rfloor}2r(r-2k+1)+ \sum_{\lfloor{N(r)}\rfloor+1}^{\lfloor{r/2}\rfloor}r^2-(r-2k+1)^2.$$ To keep the computation as simple as possible, considering that we want to find the limit $$B_r/C_r$$ for $$r\to\infty$$, we can keep only the leading terms in $$r$$ in the above expression. We can then substitute $$\lfloor{r/2}\rfloor$$ with $$r/2$$ and $$\lfloor{N(r)}\rfloor$$ with $${2-\sqrt2\over2}r$$; moreover, we can discard $$1$$ in $$r-2k+1$$. This leads to: $$B_k\approx \sum_{k=1}^{r(2-\sqrt2)/2}2r(r-2k)+ \sum_{r(2-\sqrt2)/2+1}^{r/2}r^2-(r-2k)^2= {2\sqrt2-1\over3}r^3.$$ We can repeat the same computation for $$C_k$$, obtaining: $$C_k= \sum_{k=1}^{\lfloor{r/2}\rfloor} r^2+(r-2k+1)^2\approx \sum_{k=1}^{r/2}r^2+(r-2k)^2= {2\over3}r^3.$$ Hence we obtain: $$\lim_{r\to\infty}{B_r\over C_r}= \sqrt2-{1\over2}.$$ From there, it is not difficult to show that $${\sum_{r=2}^{\infty}B_r\over \sum_{r=2}^{\infty}C_r}= \sqrt2-{1\over2},$$ because both $$B_r$$ and $$C_r$$ asymptotically grow as $$r^3$$.
The same reasoning can be repeated for the shorter leg: $$A_r=\sum_{\text{constant }r} a_{r,s}= \sum_{k=1}^{\lfloor{N(r)}\rfloor}r^2-(r-2k+1)^2+ \sum_{\lfloor{N(r)}\rfloor+1}^{\lfloor{r/2}\rfloor}2r(r-2k+1) \approx{7-4\sqrt2\over6}r^3,$$ leading to $${\sum_{r=2}^{\infty}A_r\over \sum_{r=2}^{\infty}C_r}= {7\over4}-\sqrt2.$$
• @AretinoAlso both primitive and non-primitive triplets sequenced in ascending order of $r$ and $s$ will converge to the same ratio. So this essentially settles Part 2. Part 1 in the linked question is open though. – Nilotpal Kanti Sinha Oct 26 '19 at 19:57
• What is the closed form of the corresponding ratio for the sum of the shorter sides to the sum of the hypotenuse?? Numerically it is about $\frac{7}{4} - \sqrt{2} = 0.335786$. This also proves that for a given $c$, a backward search for $a,b$ is nearly three times after than a forward search which is interesting. – Nilotpal Kanti Sinha Oct 26 '19 at 20:08