This is the final part of a question in which I have previously shown that $60$ divides $xyz$ for any primitive Pythagorean triple $(x,y,z)$. (Maybe this is relevant but I can't see why).
I am trying to find all primitive Pythagorean triples with $y=2x+1$ and $y<1000$ but I am getting nowhere. I have tried doing different things with the $x=2st, y=s^2-t^2, z=s^2+t^2$ formula but I don't seem to be getting anywhere (where I don't have to check hundreds of cases for $s$ or $t$).
On
We have $B=2A+1\quad\implies B-A=(2A+1)-A=A+1$
In primitive triples the difference between $A$ and $B$ can only be a prime number $p$ raise to any non-negative integer power where $p\equiv\pm1\mod8$. These primes are $80$ of the $168$ primes under $1000$. Under 100, the values are $1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97$. This means
$$2x+1\equiv\pm1\mod8$$ Euclid's formula is normally shown as$\quad A=m^2+n^2\quad B=2mn\quad C=m^2+n^2$
Let's try some numbers $$x=3\implies y=7\implies y-x=4$$ $$x=5\implies y=11\implies y-x=6$$ $$x=7\implies y=15\implies y-x=8$$ None of this is working so let's reverse the $A$ and $B$ functions so. $A$ is the even side. The Even side is always a multiple of $4$ so $$2(4n)+1=8n+1\equiv\pm1\mod8$$
Let's try some numbers again and refer to $\mathbb{T}$ as the set of Pythagorean triples. $$n=1\implies x=4,y=9\implies y-x=5$$ $$n=2\implies x=8,y=17\implies y-x=9$$ $$n=3\implies x=12,y=25\implies y-x=13$$ $$n=4\implies x=16,y=33\implies y-x=17$$ $$n=5\implies x=20,y=41\implies y-x=21$$ $$n=6\implies x=24,y=49\implies y-x=25$$
So far, the only candidate is. $n=4$ where $y-x=17$ but the only triples under $1000$ with this difference are
$$(45,28,53)\qquad (7,24,25)\qquad (105,88,137)\qquad (207,224,305)\qquad (555,572,797)$$
I don't know if you will find any triples that meet your criteria. It would be easier if $y=2x-1$ or $y=2x+3$.
To find triples with a given difference, you can use this formula.
$$\text{Let}\quad m=n+\sqrt{2n^2\pm p}\quad\text{where}\quad p\equiv\pm1\mod8\quad \text {to find the $m,n$ combinations for input to Euclid's formula.}$$
$\textbf{Update:}$ I did some work in a spreadsheet. There are values of x,y where $y=2x+1$ and the difference is $A+1=p^{\mathbb{Z+}}$ where $p$ is prime and $p\equiv\pm1\mod8$.
Here are the fist few values under 1000.
$$ x,y,z,\text{diff}\quad \rightarrow\quad 48,97,108.226614102078,49\quad 72,145,161.891939268143,73\quad88,177,197.668915108066,89\quad112,225,251.334438547526,113\quad128,257,287.111476607955,129\quad $$
I ran the numbers clear up past 1000 and, in none of them was $z\in\mathbb{N}$. I don't think what you are looking for exists but I do know that $y=2x-1$ has triples you might like. Let me know.
You have the right idea with what you tried. In particular, with $x = 2st$ and $y = s^2 - t^2$, you thus get
$$\begin{equation}\begin{aligned} & y = 2x + 1 \\ & s^2 - t^2 = 2(2st) + 1 \\ & s^2 - 4st - t^2 - 1 = 0 \\ & s^2 - 4t(s) + (- t^2 - 1) = 0 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, consider \eqref{eq1A} to be a quadratic in $s$. Using the quadratic formula, you then get
$$\begin{equation}\begin{aligned} s & = \frac{4t \pm \sqrt{(4t)^2 - 4(-t^2 - 1)}}{2} \\ & = \frac{4t \pm \sqrt{20t^2 + 4}}{2} \\ & = 2t \pm \sqrt{5t^2 + 1} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
For $s$ to be an integer requires that $5t^2 + 1$ be a perfect square, i.e., there needs to be an integer $u$ such that
$$5t^2 + 1 = u^2 \implies u^2 - 5t^2 = 1 \tag{3}\label{eq3A}$$
Note this is of the form of a Pell's equation. As shown in that Wikipedia article, $n = 5$ in your case, with $u$ here being $x$ there and $t$ here being $y$ there. The smallest solution of Pell equations section gives the smallest (i.e., fundamental) solution is with $n = 5$, using $u$ and $t$,
$$u_1 = 9, \; t_1 = 4 \tag{4}\label{eq4A}$$
You can easily confirm that $9^2 - 5(4^2) = 81 - 80 = 1$. Using this in \eqref{eq2A} gives, since you only want the positive value of $s$, that $s_1 = 2(4) + 9 = 17$. This then gives $x = 2(17)(4) = 136$ and $y = 2(136) + 1 = 273$.
Also, the rest of the solutions can be determined from this fundamental solution using the recursive formula given in the Additional solutions from the fundamental solution section
$$u_{k+1} = u_1 u_{k} + 5t_1 t_k \tag{5}\label{eq5A}$$ $$t_{k+1} = u_1 t_{k} + t_1 u_k \tag{6}\label{eq6A}$$
For the next set of values, this gives $u_2 = u_1(u_1) + 5(t_1)(t_1) = 9(9) + 5(4)(4) = 161$ and $t_2 = 9(4) + 4(9) = 72$. Also, you have $161^2 - 5(72^2) = 25921 - 25920 = 1$. As before, this gives $s_2 = 2(72) + 161 = 305$. This then gives $x = 2(305)(72) = 43920$ and $y = 2(43920) + 1 = 87841$. As you can see, this is already $\gt 1000$. Also, \eqref{eq5A} and \eqref{eq6A} shows the values of $u_{k+1}$ and $t_{k+1}$ are increasing, so $s_{k+1}$ will also be increasing, which means all other corresponding values of $y$ will also be larger. As such, the only set of $x$ and $y$ values which meet your requirements is $(136,273)$ mentioned earlier.