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If $a,b,c$ are Pythagorean triplets. Provided $a+b+c = s$ and $a<b<c$. We can conclude that $a\le\frac{s-3}{3}$ and $b<\frac{s-a}{2}$.

From the given conditions we can obtain: $a^{2} + b^{2} = (s-a-b)^{2}$.

I am unable to obtain those bounds for $a$ and $b$.

I tried $2a^{2} < a^{2} + b^{2} = (s-a-b)^{2} < (s-2a)^{2}$ which gives $2a^{2} > s^{2} - 2as$

I don't think this can provide me those bounds.

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    $\begingroup$ For what stands $s$ here? $\endgroup$ – Dr. Sonnhard Graubner Aug 1 '18 at 10:03
  • $\begingroup$ @Dr.SonnhardGraubner I made a mistake. $\endgroup$ – reego Aug 1 '18 at 10:06
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Are you asking for a proof of $$a\leq \dfrac{s-3}{3}\text{ and }b<\dfrac{s-a}{2}\,?$$ If so, note that $b\geq a+1$ and $c\geq b+1\geq a+2$. Therefore, $$s=a+b+c\geq a+(a+1)+(a+2)\,,$$ and the first claim is established. (Note that the only equality case for this inequality is $(a,b,c)=(3,4,5)$.)

The second claim follows from $b<c$, namely, $$s-a=b+c>b+b=2b\,.$$ We can obtain a nicer-looking (in my opinion) bound $$b\leq \dfrac{s-a-1}{2}$$ in this manner too (with the equality case $(a,b,c)=\left(2k+1,2k^2+2k,2k^2+2k+1\right)$ where $k\in\mathbb{Z}_{>0}$).

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It's known that if $(a,b,c)$ is a Pythagorean triples then there are natural $m$ and $n$ for which $m>n$ and $$\{a,b,c\}=\{m^2-n^2,2mn,m^2+n^2\}$$ Thus, $m(m+n)=500$ and number of cases.

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