Is there a way to solve for pythagorean triples with one side length equalling 1? When you have one side of a right triangle fixed, is there a trend in pythagorean triples? For instance if I fix one side of a triangle at 1 unit what will the other side have to equal to get pythagorean triplets for the first 100 values? Is there a pattern to this? 
 A: If one side (cathetus) is 1, then you cannot have a pythagorean triple because if $a$ is an integer, then $a^2 + 1$ cannot be a perfect square i.e. it cannot be equal to $b^2$ for some other integer b.   
Pythagorean triple
Note: Usually by pythagorean triple is meant 3 positive integers, not just any 3 numbers. 
A: Use rational numbers. 0.6 & 0.8 and 1.0. .36 + .64 = 1.0.  The ancients used fractions, so 3/5 and 4/5.
A: You cannot do it with Pythagorean triples but you can do it with the sum of three cubes.
$$1^3+6^3+8^3=9^3=729$$
$$1^3+71^3+138=144^3=2985984$$
$$1^3+135^3+138^3=172^3=5088448$$
$$1^3+242^3+720^3=729^3=387420489$$
$$1^3+372^3+426^3=505^3=128787625$$
$$1^3+426^3+486^3=577^3=192100033$$
$$1^3+566^3+823^3=904^3=738763264$$
A: The identity $(2mn)^2 + (m^2-n^2)^2 = (m^2+n^2)^2$ is the basis of all Pythagorean triples.  Normally, m,n are relatively prime positive integers with $m\gt n$.  To get all triples, you may need to multiply each side by a fixed positive integer.  If you allow rational sides, then 
1 : $\frac{m^2-n^2}{2mn}$ : $\frac{m^2+n^2}{2mn}$
or
$\frac{2mn}{m^2-n^2}$ : 1 : $\frac{m^2+n^2}{m^2-n^2}$
could be used as triangle sides, and similarly you could make the hypotenuse equal 1. However, there are no Pythagorean triples with integer sides and one side equal to 1.
A: Is it OK if I use $2$ instead of $1$?
THEOREM 

For all $a, b \in \mathbb Q^+$,  $2^2 + a^2 = b^2$ if and only if there exists  $\xi \in \mathbb Q^+$ such that $a = \xi - \dfrac{1}{\xi}$ and $b = \xi + \dfrac{1}{\xi}$.
PROOF
If $p, q, r \in \mathbb Z^+$
and $p^2 + q^2 = r^2$, then
there exists $u, v \in \mathbb Z^+$
such that $(p, q, r) = (2uv, u^2-v^2, u^2+v^2)$. So there are two ways to parameterize a rational right triangle so that the length of one of the sides is 2.
\begin{equation*}
   \left(2,\; \dfrac{u^2-v^2}{uv},\;\dfrac{u^2+v^2}{uv}\right)
   \text{, or }
   \left(
      \dfrac{4uv}{u^2-v^2}, \; 2, \; \dfrac{2(u^2+v^2)}{u^2-v^2}
   \right)
\end{equation*}
In the first case, we find that $\xi = \dfrac uv$. In the second case we find that $\xi = \dfrac{u+v}{u-v}$.
A: If you want sum of squares constant, then sides are 
$(\cos(t), \sin(t),1) $
Instead, if you want difference of squares constant then 
hypotenuse = $ \cosh(t) $, side = $ \sinh(t) $
The Pythagorean triplet is  $ (\cosh(t), \sinh(t),1 ) $
You can choose $t$ interval and compute them.
Either way Pythagoras thm is obeyed.
