In another post, using this formula for generating Pythagorean triples
\begin{align*} &A=(2n-1+k)^2-k^2&&=(2n-1)^2+2(2n-1)k\\ &B=2(2n-1+k)k &&=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1+k)^2+k^2 &&=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}
I got help in showing that primitive triple leg differences are limited to the set $$d\in\big\{1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, \cdots\big\}$$
and now I am working on a way to find the parameters for generating triple with specific leg differences $\,d.$
\begin{align*} |B-A|=d=2k^2-j^2\,\implies k=\sqrt{\dfrac{j^2 \pm d}{2}}\\ \end{align*}
I should be able to pick any $\,Set_n\,$ and let $\,j=(2n-1).\quad$ Then, any $\,j$-value that yields an integer indicates a triple with that leg difference in that set. For instance, if $\,d=1,\,$ \begin{align*} n=1\implies j=1\quad\implies k=\sqrt{\dfrac{1^2+1}{2}}=1\qquad F(1,1)=(3,4,5)\\ n=2\implies j=3\quad\implies k=\sqrt{\dfrac{3^2-1}{2}}=2\qquad F(2,2)=(21,20,29)\\ n=4\implies j=7\quad\implies k=\sqrt{\dfrac{7^2+1}{2}}=5\qquad F(4,5)=(119,120,169) \end{align*}
For $\,d=7,\,$ \begin{align*} n=1\implies j=1\quad\implies k=\sqrt{\dfrac{1^2+7}{2}}=2\qquad F(1,2)=(5,12,13)\\ n=2\implies j=3\quad\implies k=\sqrt{\dfrac{3^2-7}{2}}=1\qquad F(2,1)=(15,8,17)\\ n=3\implies j=7\quad\implies k=\sqrt{\dfrac{5^2+7}{2}}=4\qquad F(3,4)=(65,72,97)\\ n=3\implies j=7\quad\implies k=\sqrt{\dfrac{5^2-7}{2}}=3\qquad F(3,3)=(55,48,73)\\ \end{align*}
But then we get to $\,d=17,\,$ \begin{align*} n=1\implies j=1\quad\implies k=\sqrt{\dfrac{1^2+17}{2}}=3\qquad F(1,3)=(7,24,25)\\ n=2\implies j=3\quad\implies k=\sqrt{\bigg |\dfrac{3^2-17}{2}\bigg|}=2\qquad F(2,2)=(21,20,29)\\ \end{align*} Since $\,|20-21|\ne17,\,$ is the problem because $\,j^2<d\,$ or am I missing something?.
Answer
I will use only some of the symbols from the question in this answer. Specifically, the question starts with the following formula for Pythagorean primitives, \begin{array}{rclcl} A&=&(j+k)^2-k^2 &=&j^2+2jk,\\ B&=&2(j+k)k &=&\phantom{j^2+{}} 2jk+2k^2,\\ C&=&(j+k)^2+k^2 &=&j^2+2jk+2k^2, \end{array} where $j$ is a positive odd integer and $k$ is any positive integer relatively prime to $j$. (The "relatively prime" condition is necessary to ensure we get only primitives.)
It is then a fact that $$ B-A = 2k^2-j^2. $$ (Note: we could take the absolute value of $B-A$ on the left side, but the version without the absolute value is more specific.) From this it follows that $$ k=\sqrt{\dfrac{j^2 + B-A}{2}} $$ provided that $\dfrac{j^2 + B-A}{2}$ is a positive perfect square. In particular, $j^2 + B - A > 0$ is a necessary condition for any solution.
When $\lvert B - A\rvert = 17$, we have either $B - A = 17$ or $B - A = -17$.
Considering the case $j = 3$, $B - A = 17$, we find that $\dfrac{j^2 + B-A}{2} = 13$, which is not a perfect square.
Considering the case $j = 3$, $B - A = -17$, we find that $j^2 + B-A = -4 < 0$.
That's why there is no solution for $\lvert B - A\rvert = 17$, $j = 3$.
And that completes the answer to the question that was asked.
Discussion
The answer is not quite as short as you might like, because I took no shortcuts, because I wrote everything out explicitly, and because I considered two subcases for $\lvert B - A\rvert = 17$, $j = 3$, only one of which the question was asking about. (You already knew what happened in the other subcase.)
I looked at both sub-cases in order to avoid having to deal with any of the issues I bring up in the following discussion. In particular, I didn't have to explain how I would deal with the issue represented by the $\pm$ symbol in the question. Rather than explain how I would identify the two cases represented by the $\pm$, I simply presented both cases.
We like to maintain the conceit on this site that our questions and answers represent material that people can read and learn from. Because of that conceit, you really cannot expect to post a question and not receive a critique of the methods used in the question, even if that critique is not essential to the one thing you requested to find out.
Now consider the following two cases from the question, both of which assume that $\lvert B - A\rvert = d = 7$.
(I corrected the typos in the original post that said $j = 7$.)
According to the first line, $j = 5 \implies k = 4$. According to the second line, $j = 5 \implies k = 3$. If both lines were logically sound, then we would have $$ j = 5 \implies k = 4 \land k = 3, $$ which is false.
The resolution is that you implicitly divide the case $j = 5$ into two sub-cases, one of which yields $k = 4$ and one of which yields $k = 3$. But you buried the signal that says which sub-case is which inside a non-trivial formula that only appears on the right-hand side of the arrow. This may work for scratch notes for yourself, but if you're going to present a theory to someone else, I highly recommend being a lot more explicit about your sub-cases.
My suggestion for an easy way to make the sub-cases explicit is to work with $B - A$ directly rather than only writing it in the form $\lvert B - A\rvert$.
It's convenient to have a single letter to represent the quantity. Since $d$ is already defined in the question by $d = \lvert B - A\rvert$, I'll define $$\delta = B - A. $$
Then we have $$ k=\sqrt{\dfrac{j^2 + \delta}{2}} $$ provided that $j^2 + \delta > 0$ and that $\dfrac{j^2 + \delta}{2}$ is a perfect square.
Then the cases for $d = \lvert B - A\rvert = 7$, $j = 5$ can be tabulated as follows, because $\lvert \delta\rvert = 7$ has only the sub-cases $\delta=7$ and $\delta=-7$. \begin{array}{cc|cccc} \delta & n & j & k & F(n,k) \\ \hline 7 & 3 & 5 & \sqrt{\dfrac{5^2+7}{2}}=4 & (65,72,97) \\ -7 & 3 & 5 & \sqrt{\dfrac{5^2-7}{2}}=3 & (55,48,73) \end{array}
All I've done here is to copy the value of $\pm d$ -- either $d$ or $-d$ -- to the beginning of the line. But that makes it a lot clearer to me which sub-cases are being dealt with here.
This also makes it possible to answer the original question in just one sub-case: $$ \delta = -17,\ n = 3 \quad \implies \quad j = 5,\ j^2 + \delta = -4 < 0. $$ and therefore there is no solution in this case.
There's also a relatively trivial point of syntax. I don't want a student to come here and think that it's always a good idea to write $P \implies Q \implies R$ when what they mean to say is really "$P \implies Q$ and $Q \implies R$". I dealt with this in my discussion by making a table of values. But I suspect that a lot of math teachers using $\implies$ in this "transitive" sense on the whiteboards in front of their classes at this very moment, so this may be a lost cause.