# Is it true that for every $k\in\mathbb N$ , there exist infinitely many $n \in \mathbb N$ such that $kn+1 , (k+1)n+1$ both are perfect squares ?

Is it true that for every $k\in\mathbb N$ , there exist infinitely many $n \in \mathbb N$ such that $kn+1 , (k+1)n+1$ both are perfect squares ? What I have tried is that I have to necessarily solve $a^2-b^2=n$ for given $n$ ; but I can not proceed further . Please help . Thanks in advance

• For every $k$, one solution is $16k+8$ , maybe this helps to construct more solutions for a fixed $k$. – Peter Jan 12 '16 at 16:11
• @Peter: There's actually an infinite number of solutions. Give me a few minutes to type... :) – Tito Piezas III Jan 12 '16 at 16:29
• Another solution for every $k$ is $256k^3+384k^2+176k+24$ – Peter Jan 12 '16 at 16:32
• I know, there's a Pell equation involved. – Tito Piezas III Jan 12 '16 at 16:35
• Congratulations. I did not think of Pell ... – Peter Jan 12 '16 at 17:03

I. (Update)

The solution to,

\begin{aligned} kn+1 &= x^2\\ (k+1)n+1 &= y^2 \end{aligned}

is given by,

$$n = \frac{ -(\alpha^2 + \beta^2) + \alpha^{2(2m+1)}+\beta^{2(2m+1)} }{4k(k+1)}$$

where,

$$\alpha = \sqrt{k}+\sqrt{k+1}\\ \beta = \sqrt{k}-\sqrt{k+1}$$

For example, for $m=1,2,3,\dots$ we get,

\begin{aligned} n &= 8 + 16 k \\ n &= 24 + 176 k + 384 k^2 + 256 k^3 \\ n &= 48 + 736 k + 3968 k^2 + 9472 k^3 + 10240 k^4 + 4096 k^5 \end{aligned}

and so on.

\begin{aligned} kn+1 &= x^2\\ (k+1)n+1 &= y^2 \end{aligned}\tag1

Eliminate $n$ between them and we get the Pell-like,

$$(k+1)x^2-ky^2 = 1$$

We can get an infinite number of solutions using a transformation (discussed in this post). Let $p,\,q = 4k+1,\;4k+3$, then,

$$x = p u^2 + 2 k q u v + k (k+1) p v^2$$

$$y = q u^2 + 2 (k+1)p u v + k (k+1) q v^2$$

and $u,\color{brown}\pm v$ solve the Pell equation,

$$u^2-k(k+1)v^2 = 1$$

This has initial solution,

$$u = 2 k+1,\quad v = 2$$

and an infinite more. Thus,

\begin{aligned} n &= 8 + 16 k \\ n &= 24 + 176 k + 384 k^2 + 256 k^3 \\ n &= 48 + 736 k + 3968 k^2 + 9472 k^3 + 10240 k^4 + 4096 k^5 \\ n &= 80 + 2080 k + 20096 k^2 + 93952 k^3 + 235520 k^4 + 323584k^5 + 229376 k^6 + 65536 k^7 \end{aligned}

and so on for an infinite number of $n$ for any $k$.

• I knew there was a simpler way to express $n$. – Tito Piezas III Jan 22 '16 at 13:08