<s>[INST] Other Diophantine problems that use a Pell equation What Diophantine equations employ Pell equations in their solutions? A well-known example is the case of Pythagorean triples where the legs differ by 1, like,
$$20^2+21^2 = 29^2$$
These are completely parameterized as,
$$\big(\tfrac{x-1}{2}\big)^2+\big(\tfrac{x+1}{2}\big)^2=y^2$$
where $x^2-2y^2 =-1$. These $x,y$ also imply,
$$x^4+(y^2-1)^2 = y^4$$
$$x^4+(2y)^2 = (2y^2)^2+1$$
$$1^3+2^3+3^3+\dots+(x^2)^3 = (xy)^4$$
$$1^3+3^3+5^3+\dots+(2y-1)^3 = (xy)^2$$
Other equations that are to be solved in the integers and use Pell equations (at the least as a partial solution) are in the list below. (It is to be understood that $a,b,c,d$ are constants, while $x_i,\,y_i,\,z_i$ are unknowns.)
I. Simultaneous equations


*

*$x^2+y^2-1,\;x^2-y^2-1,\;\text{both squares}$

*$xy+1,\;xz+1,\;yz+1,\;\text{all squares}$

*$xy-1,\;xz-1,\;yz-1,\;\text{all squares}$
II. Quadratics


*

*$ax^2+bx+c = dy^2$

*$ax^2+bxy+cy^2 = d$

*$ax^2+bxy+cy^2 = dz^2+e$

*$x^2-x^2y^2+y^2=1$

*$x_1^2+x_2^2\pm1 = y^2 $

*$x_1^2+x_2^2+\dots+x_m^2\pm k = y_1^2+y_2^2+\dots+y_n^2$
III. Higher Powers


*

*$x^2+y^3 = z^4$

*$x^4+y^3 = z^2$

*$x^3+y^3+z^3+x+y+z = 0$

*$x_1^3+x_2^3+x_3^3 = 1$

*$x_1^5+x_2^5+\dots+x_7^5 = 1$

*$x_1^7+x_2^7+\dots+x_9^7 = 1$

*$x_1^4+x_2^4 = y^2+1$

*$x_1^4+x_2^4+x_3^4 = y_1^4+y_2^4+1$

*$x_1^6+x_2^6+x_3^6+x_4^6 = y_1^6+y_2^6+y_3^6+1$
Q: What other examples are not covered in the list above?
 [/INST] What two unequal numbers $m$ and $n$, letting $m<n$ are such that the sum from $m$ to $n$ is equal the product of $m$ and $n$?
$$m +(m+1) + (m+2) + ... + (n-2) + (n-1) +n = mn$$
This can actually be manipulated into
$$2(2m-1)^2-(2n-2m+1)^2=1$$
Which of course is a pell equation :
$$2\alpha^2 - \beta^2=1$$
We see that 
$$\begin{cases} \alpha=2m-1 \\ \beta=2n-(2m-1) \end{cases}$$
implying our solution set:
$$\begin{cases} m=\frac{\alpha+1}{2} \\ n=\frac{\alpha+\beta}{2} \end{cases}$$
In terms of $\alpha$ and $\beta$ as solutions of $2\alpha^2 - \beta^2=1$
$\underline{(\alpha,\beta)\to(m,n)}$
$(1,1)\to(1,1)$
$(5,7)\to(3,6)$
$(29,41)\to(15,35)$
$(169,239)\to(85,204)$
$\vdots$
and so on...
as a check verify that
$$\begin{align}
3+4+5+6 &= 3\cdot6 \\
15+16+\cdots+34+35&=15\cdot35 \\
85+86+\cdots+203+204&=85\cdot204 \\
&\vdots
\end{align}$$
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