Is the following system has any positive integer solution $(x,y,u,v)$? $$\begin{cases} x^2+y^2=u^2\\ x^2-y^2=v^2 \end{cases}$$ I can prove that any pair of these integers can be relatively prime, but I couldn't find any solution. Any hint?
Thanks in advance!
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From solutions of the Pythagorean equation we have $x = 2ab, y = a^2 - b^2$, so we must have $$(a^2 - b^2 - 2ab)(a^2 - b^2 + 2ab) = v^2$$ we have $(a^2 - b^2 - 2ab, a^2 - b^2 + 2ab) = 1$, so $$a^2 - b^2 - 2ab = (a - b)^2 - 2b^2 = x^2$$ $$a^2 - b^2 + 2ab = (a + b)^2 - 2b^2 = y^2$$ your equation has solution if and only if the Bell equations, $x^2 - 2b^2 = m^2$, $y^2 + 2b^2 = n^2$ have solutions for $(m, n)$ such that $2|m+n$, you can easily find all solutions of this equations and I think you can find some solutions this way.
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A rearrangement of variables in the "given" shows two Pythagorean triples where the odd leg of one triple matches the hypotenuse of another. \begin{equation} \qquad\qquad\qquad\qquad x^2+y^2=u^2\\ x^2-y^2=v^2\implies x^2=y^2+v^2 \end{equation} We can find matching x and matching y separatately such as
\begin{equation} x_1:x_2 \quad (17,144,145)\land(17,15,8)\quad\text{or}\quad\\ y_1:y_2 \quad (17,144,145)\land (108,144,180)\land (5183,144,5185) \end{equation}
but if two numbers of one triple match two numbers of another triple, then both the third number of both triple must also be identical. If we permit y to be different, the example of $(x_1:x_2)$ works except that $GCD(15,145)=5\land GCD(8,144)=8$. We can find such as
$$(35,12,37)\land (37,684,685)$$ but here, as in all triples, the y-term is a multiple of $4$ so it is not possible for all of the terms to be mutually prime. If you want to explore this further, the following formula permits you to find all triples that exist for a given odd number as side-C. By testing we can find the m-values in range that yield an integer k. Then we use Euclid's formula where $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor \end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$. $$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$ $$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
Just remember that side-B is always a multiple of $4$ so they can never be mutually prime.
I believe there are no (nontrivial) solutions. Note that in any solution, the numbers $v^2,x^2,u^2$ would be in arithmetic progression with common difference $y^2$. One can parametrize all three-term arithmetic progressions of squares (see for example this paper) and characterize which common differences can occur. From the paper, having a square as a common difference would correspond to nontorsion rational points on the elliptic curve $y^2=x^3-x$, which don't exist.