# 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

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

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

• $xy-1,\;xz-1,\;yz-1,\;\text{all squares}$

• $ax^2+bx+c = dy^2$

• $ax^2+bxy+cy^2 = d$

• $ax^2+bxy+cy^2 = dz^2+e$

• $x^2-x^2y^2+y^2=1$

• $x_1^2+x_2^2\pm1 = y^2$

• $x_1^2+x_2^2+\dots+x_m^2\pm k = y_1^2+y_2^2+\dots+y_n^2$

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?

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}