Inequality involving Pythagorean triplets

Question

In a Pythagorean triplet (Wikipedia), $$a=m^2-n^2\qquad b=2mn\qquad c=m^2+n^2$$ and if $L=a+b+c$, then how can we deduce that $m<\sqrt{L/2}$?

Here $m$ and $n$ are arbitrary positive integers.

Answer 1

Note that $$L=a+b+c= (m^2-n^2)+(2mn)+(m^2+n^2)=2m^2+2mn$$ Therefore $$\sqrt{\frac{L\strut}{2}}=\sqrt{m^2+mn\strut}>\sqrt{m^2\strut}=m$$ (since $m^2+mn>m^2$)

Answer 2

When you enumerate pythagorean triples, you don't want $(a, b, c)$ and $(b, a, c)$ to both come up in the enumeration as those would be duplicates. For this restrict $m$ and $n$ to ensure $a \ge b$.

$$m^2 - n^2 \ge 2mn$$ $$n/m \le \sqrt 2 - 1$$

Getting a formula for $L$ in terms of $m$ and $n$ :

$$L = m^2 - n^2 + 2mn + m^2 + n^2 = 2m^2 + 2mn$$

So an upper bound on $n$ is $n \le m(\sqrt{2} - 1)$, but this estimate in your equation overestimates the upper bound as $n \le m$. So set $n = m$:

$$L > 2m^2 + 2m^2$$

And working it out gets $m < 2^{-1}\sqrt{L}$. A better bound can be had by setting $n = m(\sqrt{2} - 1)$ and getting:

$$L \ge 2m^2 + 2m^2(\sqrt{2} - 1)$$

So $m \le 2^{-3/4}\sqrt{L}$.

Answer 3

$$L=a+b+c=2m^2+2mn=f(m,n)$$ If we solve for $n$, we get an integer only if we provide the right value of $m$ and the range of m-values is found by solving $f(m,1)$ and $f(m,m-1)$ for $m$. The smallest possible $m$ will be from for $f(m,1)$. The highest possible $m$ will be from $f(m,m-1)$. For example: \begin{equation} L_{m-1}=f(m,m-1)=2m^2+2m(m-1)=4m^2-2m\\ \implies 4m^2-2m-L=0 \implies m=\frac{\sqrt{2L+1}-1}{2} \end{equation} Now, by example $F(4,3)=(7,24,25)$ where $L=56$ and $$\frac{\sqrt{2*56+1}-1}{2}\approx 4.82\qquad \sqrt{\frac{56}{2}}\approx 5.29$$

\begin{equation} \therefore\qquad n=\frac{P-2m^2}{2m}\quad\text{for} \quad \biggl\lfloor\frac{\sqrt{4L+1}+1}{4}\biggr\rfloor\le m \le \biggl\lfloor\frac{\sqrt{2L+1}-1}{2}\biggr\rfloor<\sqrt{\frac{L}{2}} \end{equation} The lower limit ensures that $m>n$ and the upper limit insures that $n\ge1$.

$$L=56\implies \biggl\lfloor\frac{\sqrt{224+1}+1}{4}\biggr\rfloor =4\le m \le \biggl\lfloor\frac{\sqrt{112+1}-1}{2}\biggr\rfloor=4\quad\land\quad m\in\{4\}\implies n\in\{3\}$$ $$F(4,3)=(7,24,25)\qquad\quad\qquad L=(7+24+25)=56$$