You state
- Mapping $\left(x, y\right)$ to such an arithmetic progression centered at $\left(x^2+y^2\right)^2$ with difference $4xy\left(y^2-x^2\right)$ for all $\left(x,y\right)\in\mathbb{X}$, where $\mathbb{X}:=\left\{\left(x,y\right)\in\mathbb{N}^2:0<x<y\right\}$, should solve this problem,
and what you found out is that you were wrong. I will try to explain what your error was.
It took me a while to figure out that you were using the mapping
$$(x,y) \mapsto (2xy, y^2-x^2, x^2+y^2)$$
to generate Pythagorean triangles. What you have found out is that the above mapping does not generate all Pythagorean triangles.
This is the relationship between Pythagorean triples and arithmetic progressions of three perfect squares
$u^2+w^2=2v^2 \implies a^2+b^2 = c^2 \text{ where }
\left\{ \begin{array}{ccl}
a &= &\frac 12(w-u)\\
b &= &\frac 12(w+u) \\
c &= &v
\end{array}\right.$
$a^2+b^2=c^2 \implies u^2+w^2 = 2v^2 \text{ where }
\left\{ \begin{array}{ccl}
u &= & b-a\\
v &= & c \\
w &= & b+a
\end{array}\right.$
We find that
- $\quad h=v^2-u^2 = c^2 - (b-a)^2 = (a^2+b^2) - (b^2-2ab+a^2) = 2ab$
- $\quad h=w^2-v^2 = (b+a)^2 -c^2 = (b+a)^2-(b^2+a^2) = 2ab$
If we replace $a$ and $b$ with $2st$ and $t^2-s^2$ we get
$$h = 4xy(y^2-x^2) \tag{A}$$
which is just what you got; and then you found that you could not solve
$4xy(y^2-x^2)=138600$ for $x$ and $y$.
Let's go through the steps that led up to equation (A).
- $\quad (u,v,w)=(205, 425, 565)$ yields $(a,b,c) = (180, 385, 425)$
- $\quad 4xy(y^2-x^2) = 138600$
But you discovered that there is no solution to $4xy(y^2-x^2) = 138600$
We found the Pythagorean triangle that corresponds to the arithmetic progression $(205^2, 425^2, 565^2)$, so why can't we solve for $x$ and $y$?
The reason is that the mapping $(x,y) \mapsto (2xy, y^2-x^2, x^2+y^2)$ does not generate all Pythagorean right triangles!
A Pythagorean right triangle, $(a,b,c)$, is called primitive if $\gcd(a,b) =1$.
For example, $(5,12,13)$ is a primitive Pythagorean right triangle and $(9, 12, 15)$ is a Pythagorean right triangle, but it is not primitive since $\gcd(9,12) \ne 1$.
If you require that $\gcd(x,y)=1$ and $x \not \equiv y \pmod 2$ (one of $x$ and $y$ is even and the other is odd), then the mapping $(x,y) \mapsto (2xy, y^2-x^2, x^2+y^2)$ will generate all primitive Pythagorean right triangles.
If you put no restrictions on $x$ and $y$, you will generate other non primitive Pythagorean triangles but you won't generate all of them. For example, $(2xy, y^2-x^2, x^2+y^2) = (12, 9, 15)$ has no solution for $x$ and $y$.
So what you discovered is that $(180, 385, 425)$ is a non primitive Pythagorean triangle that cannot be expressed in the form $(2xy, y^2-x^2, x^2+y^2)$.
ASIDE:
The arithmetic progression of squares $(205^2, 425^2, 565^2)$ is not "primitive" either since $\gcd(205, 425, 565) = 5$. Dividing by $5$, we get the primitive triple $(u,v,w)=(41, 85, 113)$, which gives us the primitive Pythagorean triangle $(a,b,c) = (36, 77, 85)$. We then have to solve $4xy(y^2-x^2) = 85^2-41^2= 5544$. We must have $2xy = a = 36$. So $xy=18$. The possible values for $x$ and $y$ are now $(1, 18), (2, 9), (3, 6)$. We also need $y^2-x^2=b=77$. So $(x,y) = (2,9)$. No. This answer cannot be manipulated to solve the original problem.
Best Answer
Let the equation to be solved be $$ Q: v^2=6u^4-2u^2+4u+1 $$ This is an Elliptic Curve so there is a map for almost all the points to a Weierstrass form Elliptic Curve $E$ $$ E: y^2+a_1 xy+a_3 y=x^3+a_2x^2+a_4x+a_6 $$ For lucky cases, $E$ has only finitely many rational points so we can compute the inverse to check which ones are integral in $Q$. However in this case $E$ will have rank $1 \implies$ infinite rational points so this does not work directly.
There is a way to find the finitely many integer points for Weierstrass form Elliptic Curves. Seems like there is an adaptation of those methods (Elliptic Logarithms) for the quartic curve case. Magma has an implementation of it so you can use that for solving your problem.
1. Magma solver
Magma has a quartic solver exactly for your case. You can go to the online calculator and use the command (integer inputs)
for solving $$ V^2=aU^4+bU^3+cU^2+dU+e $$ where $[u,v]$ is a known integer point. Putting in
nets you the solutions $(u,v)$:
So it says that you have found all of them.
The algorithm is based on this paper. This seems to be the same methods for finding integer points on Weierstrass form Elliptic Curves but adapted for the quartic curve case.
2. Birational Equivalence to an Weierstrass form Elliptic Curve
The equation $Q$ is an Elliptic Curve so there exists a Birational Transformation to a Weierstrass form Elliptic Curve. More concretely:
The transformation can be found on page 105 of this handbook.
For $u=0$, the integer points are $(0,\pm 1)$ on $Q$. For every other integer point $(u,v),u\neq 0$ there exists a map onto a point $(x,y)$ on $E$, so we can attempt to take the inverse of all rational points on $E$ (excluding $y=0$) to see which ones map to integral $(u,v)$ on $Q$.
Edit 1: if $(u,v),u\neq 0$ maps to $y=0$ then solving the transform gives $v = -1 - 2 u + 3 u^2$. Then putting back to $Q$ we get $u=0,4$.
If $rank(E)=0$ then there are only finitely many points to check, but in this case $rank(E)=1$ so there are infinitely many rational points and I am stuck here.
The condition we need is $$ u=\frac{2(x-6)}{y} $$ is integral, where $(x,y),y\neq 0$ is a rational point on $$ y^2+4xy = x^3-6x^2-24x+144 $$ but this does not seem to be sufficient to solve it directly.
Edit 2: Group Structure of $E$
The Elliptic Curve $E$ has Mordell-Weil group structure $$ \mathbb Z \times \mathbb Z/2\mathbb Z $$ Where torsion is $T = (-6,12)$ and generator $G=(12,12)$. This can be obtained from here via translation $x=X-1$ followed by $y=Y+2X$.
Therefore all the points $P$ on $E$ can be described as $$ P = [k]G\oplus [\pm 1]T $$
i.e. addition of points on Elliptic Curve
Some of the "small points" on $E$ and its reverse map to $(u,v)$ are $$ \begin{align*} G = (12 , 12) &\mapsto(1,3)\\ T\oplus G=(0, -12) &\mapsto(1,-3)\\ [2]G = (3, -3) &\mapsto(-2,9)\\ T\oplus [2]G = (6, -24) &\mapsto(0,-1)\\ [3]G = (-4, 20) &\mapsto(-1,-1)\\ [4]G = (-15/4,-39/8) &\mapsto(4,-39)\\ T\oplus [5]G = (144/25, -12/125) &\mapsto (5,61) \end{align*} $$ These generated the solutions we are looking for.