Proving there are infinitely many pairs of square-full consecutive integers

So the question goes like this :

A positive integer $n$ is called square-full if for every prime $p$, $p \, | \, n$ implies $p^2 \, | \, n$, i.e. every prime power factor of $n$ occurs at least at the second power. This is equivalent to saying that there is no prime $p$ such that $p \, \| \, n$.

Show that there are infinitely many consecutive pairs of square-full numbers, i.e. infinitely many $n$ such that $n$ and $n+1$ are square-full.

The context of the question highly suggests that this can be proved by induction and it is the solution that I am looking for. It is not a homework, just something that caught my attention because I couldn’t find any natural answer to it… I hope I’m not missing something trivial!

I’ve tried assuming $n_1, n_2, \dots, n_k$ were the first $n_k$ integers such that $n_i$ and $n_i+1$ are square-full, and then generate some polynomial in those $2k$ integers to get a new couple $n_{k+1}$, $n_{k+1} + 1$ but after a few natural tries I got nothing good out of it. Any ideas?

Solutions Collecting From Web of "Proving there are infinitely many pairs of square-full consecutive integers"

Any product of square-full numbers is square-full.

$4n(n+1) $ is square-full if n, n+1 are.
$4n(n+1)+1=4n^{2}+4n+1=(2n+1)^{2}$

list of square-full numbers at http://oeis.org/A060355

Suppose that $a_0=1$, $a_1=3$, $b_0=0$, and $b_1=2$. Let $a_n=6a_{n-1}-a_{n-2}$ and $b_n=6b_{n-1}-b_{n-2}$. Then $a_n^2-2b_n^2=1$ and $a_na_{n-1}-2b_nb_{n-1}=3$.

This means that $a_n^2$ and $2b_n^2$ are adjacent integers, and since $b_n$ is even both are obviously square-full for $n>0$.

Inductive Verification:
$$
\begin{align}
a_n^2-2b_n^2
&=(6a_{n-1}-a_{n-2})^2-2(6b_{n-1}-b_{n-2})^2\\
&=(36a_{n-1}^2-12a_{n-1}a_{n-2}+a_{n-2}^2)-2(36b_{n-1}^2-12b_{n-1}b_{n-2}+b_{n-2}^2)\\
&=36(a_{n-1}^2-2b_{n-1}^2)+(a_{n-2}^2-2b_{n-2}^2)-12(a_{n-1}a_{n-2}-2b_{n-1}b_{n-2})\\
&=36+1-36\\
&=1
\end{align}
$$
and
$$
\begin{align}
a_na_{n-1}-2b_nb_{n-1}
&=(6a_{n-1}-a_{n-2})a_{n-1}-2(6b_{n-1}-b_{n-2})b_{n-1}\\
&=(6a_{n-1}^2-a_{n-1}a_{n-2})-2(6b_{n-1}^2-b_{n-1}b_{n-2})\\
&=6(a_{n-1}^2-2b_{n-1}^2)-(a_{n-1}a_{n-2}-2b_{n-1}b_{n-2})\\
&=6-3\\
&=3
\end{align}
$$