Bounds in Pythagorean Triples Let's take a look at this simple task (Pythagorean Triples):
Calculate $A$ and $B$ such that $A^2 + B^2 = C^2$.
$C$ is given.
Is there any way to find an upper bound for $A$, $B$, $A^2$, and $B^2$?
The upper bound will be a function of $C$.
$0 < A < f_1(C)$
$0 < B < f_1(C)$
$0 < A^2 < f_2(C)$
$0 < B^2 < f_2(C)$
$f_1(C) = ?$
$f_2(C) = ?$
Any ideas?
 A: Note that $3,4,5$ is a triple and $n^2-1,2n, n^2+1$ is also a triple. 
Your upperbound must satisfy 
$$f(n^2+1) > n^2-1$$
So if you are looking for a "nice function" (Polynomial), the best upperbound must satisfy $f(C) \geq c-2$ infinitelly often, and you can  make it $C-1$. To replace $C-2$ by $C-1$ you just observe that $k,n,n+1$ is a solution whenever $k^2=2n+1$, that is whenever $n=\frac{m^2-1}{2}$ for $m$ an odd integer. 
Moreover, these functions yield $\leq$ not strict inequalities. If you want strict inequalities, there is no best polynomial upperbound. You might want your function to only take integer values.
If by function you mean any function, then the answer is simple: your problem is the definition of your function. And again, since all your variables are integers, there is no best upperbound to satisfy your strict inequalities, unless again you restrict the codomain to integers.
A: There are an infinite number of pythagorean triples for which $B + 1 = C$.  Let $A = 2n+1$.  Let $B = 2n^2 + 2n$ and $C = B + 1 = 2n^2 + 2n + 1$.
$$\begin{align}
\\ A^2 + B^2 &= (2n+1)^2 + (2n^2 + 2n)^2
\\ & = (4n^2 + 4n + 1) + (4n^4 + 8n^3 + 4n^2)
\\ & = 4n^4 + 8n^3 + 8n^2 + 4n + 1
\\ & = (2n^2 + 2n + 1)^2
\\ & = C^2
\end{align}
$$
