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 
 A: 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 $k$.
