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.

  • $\begingroup$ @Aretino Here is a similar problem with a different sequencing order of the triplets $\endgroup$ – Nilotpal Kanti Sinha Oct 25 '19 at 10:27
  • $\begingroup$ 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. $\endgroup$ – Aretino Oct 25 '19 at 16:08
  • $\begingroup$ @Aretino Added that now. Regarding order, we just need to generate all triplets for $r \le x$ for some $x$ and make $x \to \infty$ $\endgroup$ – Nilotpal Kanti Sinha Oct 25 '19 at 16:16
  • $\begingroup$ You should add that all sides are integer numbers. $\endgroup$ – Ripi2 Oct 25 '19 at 16:55
  • $\begingroup$ @Ripi2 That is already in the word Pythagorean triplet because by definition all numbers in a Pythagorean triplet are natural numbers $\endgroup$ – 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: $$ b_{r,s}=\max(r^2-s^2, 2rs)= \cases{ 2r(r-2k+1) & for $1\le k<N(r)$,\\ r^2-(r-2k+1)^2 & for $N(r)<k\le\lfloor{r/2}\rfloor$,\\ } $$ 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 $$ {\sum_{r=2}^{\infty}A_r\over \sum_{r=2}^{\infty}C_r}= {7\over4}-\sqrt2. $$

  • $\begingroup$ @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. $\endgroup$ – Nilotpal Kanti Sinha Oct 26 '19 at 19:57
  • $\begingroup$ 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. $\endgroup$ – Nilotpal Kanti Sinha Oct 26 '19 at 20:08
  • $\begingroup$ @NilotpalKantiSinha Yes, that's exactly the value one finds, see my edit above. $\endgroup$ – Aretino Oct 26 '19 at 20:48

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