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 ?

Question

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

Answer 1

The answer is Yes.

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.

II. (Old answer)

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

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 $k$.