Let $w<x$ be integers with $x^2-w^2$ beeing a square. If no $y$ exists with $y^2-x^2$, $y^2-w^2$ being squares, can we show no $y$ exists for $nw<nx$?

Question

There exist vast pairs $(w,x)$, $w<x$ with $x^2-w^2=\square_1$ beeing a square (also known as Pythagorean Triples). I am trying to distinguish two "classes" of such pairs:

• Those pairs for which no $y$ exists that satisfies $y^2-x^2=\square_2$, $y^2-w^2=\square_3$ being again perfect squares. In the plot below these pairs are colored red. Some examples of such "useless pairs" are $(4,5),(8,10),(24,26)$.
• Those pairs for which a $y$ exists satisfying that condition. These pairs are the green dots in the chart below. Some examples of such "promising pairs" are $(153,185),(264,520),(264,561)$.

When taking a look at this plot I came to the question: In the case we know that for a pair $(w,x)$ no such $y$ exists with $y^2-x^2$, $y^2-w^2$ being squares, can we formally show that no $y$ exist for any pair $(n\cdot w,n\cdot x)$, $n\in\mathbb{N}$?

Background and ideas: With Arty we implement and run a program searching for triples $(w,x,y)$ and even quadruples $(w,x,y,z)$ whose difference of squared elements produce again squares.

We could speed up our search by being able to distiguish "useless pairs" (those pairs for which we will not find an integer $y$ completing the pair to a triple whose difference of squared elements produce again squares) from "promising pairs" (those ones for which an integer $y$ exists).

Another idea was to use a DecisionTreeClassifier who tries to distinguish between "useless pairs" (red dots in the plot) and "promising pairs" (green dots) based on the slope $\frac{x}{w}$. This classifier achieved an accuracy of $0.9065$ (thee this notebook at GitHub).

Answer 1

For $\space w^2 + x^2 =y^2\space$ to work, when $\space (w,x,y)\space$ are divided by $\space g=GCD(w,x,y),\space$ then the first two terms must be of opposite parity so that the third can be odd. If both are the same parity, then the third term will be even, no matter what the multiplier. Since $\space GCD(153,185)=1,\space$ and the terms are both odd, no $\space y$-value can exist for them no matter what the multiplier.

In another example, your $\space (264,520)=8\times(33,65)\space$ so the same is true; neither can yield a $\space y$-value that, when multiplied by the $\space GCD.\space$ completes a valid triple.

We can find $\space 3 \space$ squares whose differences are all squares but not $\space 4\space $ as shown for your previous questions here and here.