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

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 $u+v$.
