Number Theory involving Pythagorean triplets

Question

With $n \neq 4$ find all natural numbers $(n,k)$ such that $n^2 + (n-1)^2 = k^2$

This is part of a problem I am working on. I am trying to find out whether $n=4$ is the only answer or there exist many more answers.

I tried using the theorem for primitive Pythagorean triplets setting $n=u^2-v^2$ and $n-1=2uv$ where $(u,v)=1$. However I couldn't get anywhere after this

Please help on how to go forward I'm stuck!

Edit- Further, is there a way of generating values on $n$ ?

Answer 1

You have $u^2-v^2=1+2uv$ and $(u-v)^2-2v^2=1$. This is a Pell equation. You can use this to generate all possible values of $n$.

Answer 2

You can rewrite $$n^2+(n−1)^2=k^2$$ to get the negative Pell equation: $$m^2 - 2k^2 = -1$$ where $m=2n-1$.

If I'm not mistaken, the solutions for $m$ are $1$, $7$, $41$, $239$, $1393$, $8119$, $47321$, $275807$, $1607521$, $9369319$, ...

This is given by the formula:

$$m_i = \frac{1+\sqrt{2}}{2} (3+2\sqrt{2})^i + \frac{-1+\sqrt{2}}{2} (3-2\sqrt{2})^i$$

That second term is very small, so you can just use the first term and round it up to an integer.

$$m_i = \lceil\frac{1+\sqrt{2}}{2} (3+2\sqrt{2})^i\rceil$$

Answer 3

As I recall, this formula generates all such triples: For the triple $(a,b,c)=(4,3,5)$, we have that $m=2, n=1$, so that $a=4=2mn$ and $b=3 =m^2-n^2.$ Define the sequences recursively $m_1 = 2$, $n_1 = 1$ and then $m_{j+1} = 2m_j+n_j$ and $n_{j+1}=m_j.$

Then $m_2 = 5, n_2 =2,$ giving the triple $(20,21,29)$

Then $m_3 = 12, n_3 =5$, giving the triple $(119,120,169)$

Etc.

Answer 4

There are infinite solutions though the $20^{th}$ and subsequent triples exceed $15$ digits. We begin with a "solved" problem by effectively reversing $A$ and $B$ in this equation:

$$A^2+B^2=C^2\quad\text{where}\quad B=A+1$$ With a starting "seed" of $ T_0=(A_0,B_0,C_0) = (0,0,1)\quad$

\begin{equation}T_{n+1}:\quad A_{n+1}=3A_n+2C_n+1\quad B_{n+1}=3A_n+2C_n+2 \quad C_{n+1}=4A_n+3C_n+2\end{equation}

$$T_1=(3,4,5)\quad T_2=(20,21,29)\quad T_3=(119,120,168)\quad T_4=(696,697,985)\quad\textbf{ ...}$$

Another means of generating them sequentially (with $A>B$ and $B>A$ alt4rnating) is to find the $(m,k)$-values needed to use Euclid's formula:

$$ A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2$$

\begin{equation} \quad m=k+\sqrt{2k^2+(-1)^k} \end{equation} \begin{align*} k=1\quad &\implies m=\bigg(1+\sqrt{2(1)^2+(-1)^1}\space\bigg)=2\quad & F(2,1)=(3,4,5)\\ k=2\quad &\implies m=\bigg(2+\sqrt{2(2)^2+(-1)^2}\space\bigg)=5\quad & F(5,2)=(21,20,29)\\ k=5\quad &\implies m=\bigg(5+\sqrt{2(5)^2+(-1)^5}\space\bigg)=12\quad & F(12,5)=(119,120,169)\\ k=12\quad &\implies m=\bigg(12+\sqrt{2(12)^2+(-1)^{12}}\space\bigg)=29\quad & F(29,12)=(697,696,985) \end{align*}

These are Pell numbers and a way to generate the $n^{th}\space (m,k)\space$ pair directly is

\begin{equation} m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad \qquad k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}} \end{equation}

\begin{align*} &\frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 \space & \frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1 \quad& F(2,1)=(3,4,5)\\ & \frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 \space & \frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2 \quad& F(5,2)=(21,20,29)\\ & \frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 \space & \frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5 \quad& F(12,5)=(119,120,169)\\ &\frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 \space & \frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12 \quad& F(29,12)=(697,696,985) \end{align*}

To answer the original question $\quad n\in\{4,21,120, 697, \cdots\}$