Pythagorean Triple Inequality When finding the Pythagorean triple where $a+b+c=1000$,
Wolfram alpha shows me that $a< -500\left(\sqrt{2} - 2\right)$
When I input $a^2+b^2=c^2, a<b<c$ and $a+b+c=1000$
How does wolfram arrive at that inequality:
$a< -500\left(\sqrt{2} - 2\right)$
Here is the link: Wolfram
 A: Note:  seeing the edited form of the question, the following isn't really on point.  It addresses an efficient algorithm for finding the triple, it does not speak to whatever Wolfram Alpha might be doing.
Playing with the code, the following emerges:  Suppose $$a^2+b^2=c^2\;\;\;\&\;\;\;a+b+c=S$$
We remark that $S^2-2Sb=2a^2+2ac=2a(a+c)=2a(S-b)$  Whence:  $$a=\frac {S^2-2Sb}{2(S-b)}$$
Inelegant, perhaps, but it makes the computer search very efficient.
A: For the inequality, we observe that since $b > a$, we must have
$$
c^2 = a^2+b^2 > 2a^2
$$
or
$$
c > a\sqrt{2}
$$
Thus
$$
1000 = a+b+c > a+a+a\sqrt{2} = a(2+\sqrt{2})
$$
or
$$
a < \frac{1000}{2+\sqrt{2}} = \frac{1000(2-\sqrt{2})}{2^2-(\sqrt{2})^2}
  = 500(2-\sqrt{2})
$$
As regards the original problem: Primitive Pythagorean triples can be obtained using the well-known schema
$$
a, b = u^2-v^2, 2uv
$$
in some order, with $u > v$ both integers, and
$$
c = u^2+v^2
$$
Note that in this case,
$$
a+b+c = 2u^2+2uv = 2u(u+v)
$$
In order for $ka+kb+kc = 1000$ for some integer $k$, we need $u(u+v) \mid 500$.  For example, with $u = 20, v = 5$, we obtain $a = 2uv = 200, b = u^2-v^2 = 375, c = u^2+v^2 = 425$.  Note that $a+b+c = 200+375+425 = 1000$, and $a^2+b^2 = 40000+140625 = 180625 = c^2$.
That this is the only solution can be demonstrated by noting first that $500 = 2^2 \times 5^3$, and observing that for $u$ and $u+v$, we need two disjoint subsets of factors whose separate products differ by less than a factor of $2$.*  This only happens for the above case with $20$ and $25$ (yielding $u = 20, v = 5$), and also with $4$ and $5$ (yielding $u = 4, v = 1$, and requiring us to scale the resulting triple by a factor of $25$).  Since these two pairs are in direct proportion to each other, they produce the same unique solution.
*ETA: Only if we require $a > 0$.  If $a$ can be less than $0$, then we only need disjoint factor subsets $u$ and $u+v$.
A: Let $(A+B+C)=P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mn$
$$\text{We solve for  }\\ n=\frac{P-2m^2}{2m}\text{ where }\biggl\lceil\frac{\sqrt{P}}{2}\space\biggr\rceil\le m\le\biggl\lfloor\sqrt\frac{P}{2}\biggr\rfloor$$
Given $$P=1000\implies m_{min}=\biggl\lceil\frac{\sqrt{1000}}{2}\space\biggr\rceil=16, \quad m_{max}= \biggl\lfloor\sqrt{\frac{1000}{2}} \biggr\rfloor =22\\
\text{ and
 we test }16\le m\le 22 \quad \text{to see which yield }\quad
n \in\mathbb{N}$$
We find only one where $f(20,5)=(375,200,425)$. Note that $A$ and $B$ are reversible if $a<b<c$ is a concern. In any case it's a multiple: $25(15,8,17)$
