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\begin{center}
\vskip 1cm{\LARGE\bf Powerful Values of Quadratic Polynomials
}
\vskip 1cm
\large
Jean-Marie De Koninck and Nicolas Doyon \\
D\'epartment de Math\'ematiques \\
Universit\'e Laval \\
Qu\'ebec, PQ G1V 0A6 \\
Canada \\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca} \\
\href{mailto:nicolas.doyon.1@ulaval.ca}{\tt nicolas.doyon.1@ulaval.ca} \\
\ \\
Florian~Luca \\
Instituto de Matem{\'a}ticas\\
Universidad Nacional Aut\'onoma de M{\'e}xico \\
C. P. 58180\\
Morelia, Michoac{\'a}n \\
M{\'e}xico \\
\href{mailto:fluca@matmor.unam.mx}{\tt fluca@matmor.unam.mx}\\
\end{center}

\vskip .2 in

\begin{abstract}
We study the set of those integers $k$ such that $n^2+k$ is powerful for
infinitely
many positive integers $n$. We prove that most integers $k$ have this
property.
\end{abstract}

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\section{Introduction}

Given an arbitrary integer $k\neq 0$, Mollin and Walsh
\cite{kn:mollinwalsh} have shown that there exist infinitely many ways
of writing $k$ as a difference of two nonsquare powerful numbers. A
positive integer $n$ is said to be {\it powerful} if $p^2 \divides n$ for each
prime divisor $p$ of $n$. For instance, 17 has two such representations
below $10^9$, namely $17=125-108$ and $17=173881809 - 173881792$. But
what about representations $17=P-n^2$, where $P$ is a powerful
number?  It turns out that there are no such representation with
$P&lt;10^9$. However, in view of Theorem 1 (below) we believe that
infinitely many such representations should exist, even though the
smallest is probably  very large (see Table 1 in Section 3). In
general, identifying all those integers $k\neq 0$ such that $n^2+k$ is
a powerful number for infinitely many positive integers $n$ seems to be
a very difficult problem.  Indeed, already showing that one such $n$
exists is not obvious.

Now, for a given $k\neq 0$, if one
can find an integer $n_0$ such that $n_0^2+k$ is a powerful number which is
not a perfect square, that is if
\begin{equation}
n_0^2+k = m_0^2 b^3, \label{eq:ex1}
\end{equation}
for some integer $m_0$, with $b&gt;1$  squarefree,
then (\ref{eq:ex1}) can be written as
\begin{equation}
n_0^2 - D m_0^2 = -k , \label{eq:ex2}
\end{equation}
where $D&gt;1$ is a cube but not a square.
Now since $x^2-Dy^2=-k$ is a generalized Pell equation  with a given solution 
(see, for instance, Robertson \cite{rob}), it must
have infinitely many solutions, thus providing infinitely
many $n$'s for
which $n^2+k$ is powerful.
However, given an arbitrary integer $k$, finding a minimal solution
$(n_0,m_0)$ of (\ref{eq:ex2}) for an appropriate $D$ is not easily achieved.

In this note, we use a different approach and show that for almost all
integers $k$, there exist infinitely many positive integers $n$ such that
$n^2+k$ is powerful. In fact, we prove the following result.

\begin{theorem}
\label{thm:main}
For any positive real number $x$, let ${\cal P}(x)$ be the set of integers
$k$ with $|k|\le x$ such that
$n^2+k$ is powerful for infinitely many positive integers $n$.
Then $2x-\#{\cal P}(x)=o(x)$.
\end {theorem}


Throughout this paper we use the Landau symbols $O$ and $o$ as well as the
Vinogradov symbols $\gg$ and $\ll$ with their usual meaning. We let
$\log_1 x=\max\{1,\log x\}$, where $\log$ stands for the
natural logarithm, and for $i&gt;1$ we define $\log_i x=\log_1\(\log_{i-1} x\)$.
When $i=1$ we omit the subscript and thus understand that all the logarithms
that will appear are $\ge 1$. For a positive integer $n$ we write $\phi(n)$
for the Euler function of $n$.

\medskip

\section{The Proof}

For the proof, we first give a sufficient algebraic criterion on
$k$ which insures that $n^2+k$ is powerful for infinitely many $n$. We then
show that most integers $k$ satisfy this condition.

\medskip

We shall prove this only when $k&gt;0$, but the argument extends without
any major modification to the case $k&lt;0$.

\begin{proposition}
\label{prop:alg}
Assume that there exist positive integers $y_1$ and $d \divides ky_1^2+1$ such that
$$
u=\frac{1}{2y_1}\left(\frac{ky_1^2+1}{d}-d\right)
$$
is a positive integer coprime to $k$. Let $D=u^2+k$ and assume further that
$y_1$ is coprime to $D$. Then there exist infinitely many positive integers
$n$ such that $n^2+k$ is powerful.
\end{proposition}

\begin{proof} Since $u$ is an integer, it follows that $d$ and $(ky_1^2+1)/d$
are integers of the same parity. Put
$$
x_1=\frac{1}{2}\left(\frac{ky_1^2+1}{d}+d\right).
$$
One checks immediately that $x_1^2-(uy_1)^2=ky_1^2+1$, which can be rewritten
as
\begin{equation} \label{eq:ajouteq}
x_1^2-Dy_1^2=1.
\end{equation}
Define the sequences $(x_m)_{m\ge 1}$ and
$(y_m)_{m\ge 1}$ as
$$
x_m+y_m{\sqrt {D}}=(x_1+y_1{\sqrt {D}})^m
$$
for all $m\ge 1$. Then, for all $m\ge 1$,
\begin{eqnarray}
\label{eq:pell1}
x_m^2-Dy_m^2&amp; = &amp; (x_m +  y_m \sqrt D) (x_m -  y_m \sqrt D) \\ \nonumber
&amp; = &amp; (x_1+y_1 \sqrt D)^m  (x_1- y_1 \sqrt D)^m \\ \nonumber
&amp; = &amp; (x_1^2 - D y_1^2 )^m = 1
\end{eqnarray}

We now search for positive integers $n$ such that $n^2+k=D\ell^2$ holds with
some positive integer $\ell$ such that $D\divides \ell$. It is clear that
such numbers $n$ have the property that $n^2+k$ is powerful.
We rewrite this equation as
\begin{equation}
\label{eq:pell2}
n^2-D\ell^2=-k.
\end{equation}
Noting that $u^2-D\cdot 1^2=-k$ and using \eqref{eq:pell1}, if
$$
n+{\sqrt {D}}\ell =(u+{\sqrt {D}})(x_m+{\sqrt {D}}y_m),
$$
one checks by multiplying each side  by its conjugate that the pair $(n,\ell)$ satisfies \eqref{eq:pell2}. Expanding we get
$$
n=ux_m+Dy_m\qquad {\text{\rm and}}\qquad \ell=uy_m+x_m.
$$
It suffices to argue that there exist infinitely many $m$ such that
$D\divides \ell$. Since
$$
x_m=\frac{1}{2}\left((x_1+y_1{\sqrt {D}})^m+(x_1-y_1{\sqrt {D}})^m\right)
\equiv x_1^m \pmod D,
$$
and
$$
y_m=\frac{1}{2{\sqrt {D}}}\left((x_1+y_1{\sqrt {D}})^m-
(x_1-y_1{\sqrt {D}})^m\right)\equiv mx_1^{m-1}y_1\pmod D,
$$
the relation $D\divides (uy_m+x_m)$ is equivalent to $D \divides x_1^{m-1}(umy_1+x_1)$.
Since $D$ and $x_1$ are coprime (in light of (\ref{eq:ajouteq})), the above divisibility relation
holds if and only if $muy_1\equiv -x_1 \pmod D$. Since both
$u$ and $y_1$ are coprime to $D$, it follows that their product is  invertible
modulo $D$. Hence, if $m\equiv -x_1(uy_1)^{-1} \pmod D$, then
$$
n=ux_m+Dy_m
$$
has the property that $n^2+k$ is powerful. This completes the proof of the
proposition.
\end{proof}

It now remains to show that for most positive integers $k$ one can choose
integers $y_1$ and $d$ such that the conditions from Proposition
\ref{prop:alg} are fulfilled. It is clear that for the purpose of making
$n^2+k$ powerful, we may assume that $k$ is squarefree. Indeed,  if
$p^2\divides k$, we may then take $n=pn'$ and note that
$$
n^2+k=p^2(n'^2+(k/p^2)),
$$
so we may replace $k$ by $k/p^2$.

\begin{theorem}
\label{thm:anal}
The set of squarefree positive integers $k$ for which there exist positive integers $y_1$ and
$d\divides ky_1^2+1$ such that the conditions of Proposition \ref{prop:alg} are
satisfied is of density 1.
\end {theorem}

\begin{proof}
We let $x$ be a large positive real number and we assume that
$k\le x$ is a positive integer. We choose $y_1=12$.

\medskip

The number $d$ will always be a prime number in a certain arithmetical
progression modulo $144$, as follows. If $\gcd(k,6)=1$, we then take
$d\equiv 1\pmod {144}$. If $\gcd(k,6)=2$, then $d\equiv 91\pmod {144}$.
If $\gcd(k,6)=3$, we put $d\equiv 65\pmod {144}$, and finally if
$6\divides k$, we then put $d\equiv 11\pmod {144}$.

\medskip

We first show that $y_1$ and $D$ are coprime, from which it will follow  that
$6$ and $D$ are coprime.

If $\gcd(k,6)=1$, then since $d\equiv 1 \pmod {144}$, we get that
$(144k+1)/d\equiv 1\pmod {144}$. Hence, $(144k+1)/d-d\equiv 0\pmod {144}$,
which shows that $6\divides u$. Since $k$ is coprime to $6$, we get that
$6$ is coprime to $D$.

If $\gcd(k,6)=2$, then $d\equiv 91\pmod {144}$. In particular,
$d\equiv 11\pmod {16}$ and $d\equiv 1\pmod 9$. Hence, $(144k+1)/d\equiv
3\pmod {16}$ and $(144k+1)/d\equiv 1\pmod 9$. Thus,
$(144k+1)/d-d$ is congruent to $8$ modulo $16$ and to $0$ modulo $9$. Hence,
$u$ is an odd multiple of $3$. Since $2$ divides $k$ but $3$ doesn't, we get
that $6$ is coprime to $D$.

It is easily seen that the other two cases, namely $\gcd(k,6)=3$ and
$\gcd(k,6)=6$,  can be treated similarly.

Moreover, since $\mbox{gcd}(u^2+k,12)=\mbox{gcd}(D,y)=1$  there is no prime
$p\in\{2,3\}$ such that $p\divides \gcd(k,u)$.

\medskip

Thus, it remains to show that for all positive integers $k\le x$
except $o(x)$ of them such a prime $d$ can be
chosen in such a way that there is no prime $p&gt;3$ dividing both $u$ and $k$.
Note that if $p&gt;3$ divides both $u$ and $k$, then $(144k+1)/d\equiv d\pmod p$,
so that $144k+1\equiv d^2\pmod p$, in which case $d^2\equiv 1\pmod p$.
Thus, $d\equiv \pm 1\pmod p$.
We can reverse the argument to show that if $d\equiv \pm 1 \pmod p$, then $p
\divides 2y_1 u$ and $p \divides k$.
Since $p&gt;3$ and the largest prime factor of
$y_1$ is $3$, the condition $d\equiv \pm 1 \pmod p$ guarantees that
$p \divides \gcd(u,k)$.

\medskip

For coprime positive integers $a,~b$ we write
$$
S(x;a,b)=\sum_{p\equiv \fmod{a} {b}}\frac{1}{p}-\frac{\log_2 x}{\phi(b)}.
$$
A result of Pomerance (see Theorem 1 and Remark 1 in \cite{Pom})
shows that, uniformly for all $a&lt;b\le x$,
$$
S(x;a,b)=\frac{1}{p(a,b)}+O\left(\frac{\log 2b}{\phi(b)}\right),
$$
where $p(a,b)$ is the smallest prime number in the arithmetical
progression $a\pmod b$.

\medskip

Let $\omega(k;a,b)$ be the number
of prime factors $p$ of $k$ which are congruent to $a\pmod b$.
We let $b=144$, $a=1,91,65,11$ according to whether
$\gcd(k,6)=1,2,3,6$, respectively. By a classical result of
Tur\'an \cite{Tur}, the estimate
$$
\omega(144k+1;a,b)=\frac{\log_2 x}{\phi(b)}+O\left(\left(\frac{\log_2 x}{\phi(b)}
\right)^{2/3}\right)
$$
holds
for all $k\le x$, with at most
$$
O\left(\frac{x}{(\log_2 x)^{1/6}}\right)=o(x)
$$
exceptions. From now on, we work only with such positive integers
$k$. Note that $\omega(144k+1;a,b)$ gives the number of admissible values
for $d$.

\medskip

We now put $y=\log_2 x$, $z=x^{1/3}$ and
show that the number of such $k\le x$ for which there exists a prime factor
$p\in  [y,z]$ dividing both $u$ and $k$ is $o(x)$. Indeed, let us fix
$p$ and $d$. Then $k\equiv 0\pmod p$ and $144k+1\equiv 0\pmod d$.
This puts $k\le x$ into a certain arithmetical progression modulo
$pd$.


\medskip

Assume first that $pd\le x$. Clearly, the number of such
positive integers $k\le x$ is $\le x/(pd)+1\ll x/(pd)$.
Summing up this inequality for all  $p\ge y$
and all $d\equiv \pm 1\pmod p$,
we get that the number of such numbers $k$ is
$$
 \ll   x \sum_{p\ge y}\frac{1}{p}
\sum_{\substack{d\le x \\ d\, \equiv \, \pm  \fmod{1} {p}}} \frac{1}{d}
 \ll  x\log_2 x\sum_{p\ge y}\frac{1}{p^2}
 \ll  \frac{x\log_2 x}{y\log y}=o(x).$$ 

We now look at those positive integers
$k\le x$ such that $pd&gt;x$. Write $144k+1=dm$, and note that
$m\le 288 x/d&lt;288 p\ll p$.
Since $p \divides k$ and $d\equiv \pm 1\pmod p$, we get
that $m\equiv \pm 1\pmod p$. Fix $m$. Then $k/p$ is in a certain residue
class modulo $m$ depending on $p$. Write $k/p=v+m\ell$. Then
$$
dm=144p(k/p)+1=(144pv+1)+144pm\ell,
$$
so that
$$
d=w+144p\ell,
$$
where $w=(144pv+1)/m$. Furthermore, $d\le 288 x/m$. Hence, by a result of
Montgomery and Vaughan \cite{MV}, the number of such primes $d$ does not
exceed
\begin{equation} \label{eq:spe}
\frac{4\cdot 144 x}{m\phi(144p)\log(288 x/(mp))}\ll \frac{x}{mp\log x},
\end{equation}
where we used the fact that
$m\le 288p  \le 288 x^{1/3}$, and therefore
$$
\frac{288x}{mp}\gg x^{1/3}.
$$
Summing up inequality (\ref{eq:spe})  over all the possible values of
$m\le 288 p\le 288 x^{1/3}$, and then  afterwards over all  $p\in [y,z]$,
we get that the number of such $k$'s is
$$ \ll  \frac{x}{\log x}
\sum_{y\le p}\frac{1}{p}\sum_{\substack{m\le x \\
m \,\equiv \, \pm \fmod{1} {p}}}\frac{1}{m}
 \ll  x\sum_{y\le p}\frac{1}{p^2}\ll \frac{x}{y\log y}=o(x).$$

\medskip

From now on, we consider only those $k$ such that if $d\equiv a\pmod b$
is a prime factor of $144k+1$, with the pair $(a,b)$ being the appropriate
one depending on the value of $\gcd(k,6)$, then there exists
a prime $p\in [5,y]\cup [x^{1/3},x]$ such that $p \divides \gcd(k,u)$.
First observe that $k$ has at most 3 prime factors in $[x^{1/3},x]$.

Moreover, for each prime $p&gt;x^{1/3}$, there are at most
$3$ values of $d$ such that $d\equiv \pm 1\pmod p$. Indeed, if there
were 4 or more, let $d_1,~d_2$,
$d_3$ and $d_4$ be 4 of them. We would then have
$$
144x+1\ge 144k+1\ge d_1d_2d_3d_4\ge (x^{1/3}-1)^4,
$$
which
is impossible for large $x$. Hence, there are at most
$9$ values of $d$ which might be congruent to
$\pm 1$ modulo some prime factor $p&gt;x^{1/3}$ of $k$.
Since we have $(1+o(1)){\displaystyle{\frac{\log_2 x}{\phi(b)}}}$ such
primes $d$, it follows that
we also have  $(1+o(1)){\displaystyle{\frac{\log_2 x}{\phi(b)}}}$
such prime factors $d$ of $144k+1$
with the property that each of them is congruent to $\pm 1$ modulo a prime
factor $p$ of $k$ in the interval $[5,y]$. We apply Tur\'an's inequality
from \cite{Tur}
again to conclude that all $k\le x$ have at most $1.5\log_2 y&lt;2\log_4 x$
prime factors
$p&lt;y$ with at most  $o(x)$ exceptions.

\medskip

We now write
$$
M=\prod_{5\le p&lt;(\log_3 x)/2} p,
$$
and look at those $d$ such that
$d\equiv 2\pmod M$.  Note that such $d$ are in a certain arithmetical
progression $A \pmod B$, where
$B=bM=(\log_2 x)^{1/2+o(1)}$. We apply again the results from \cite{Pom} and
\cite{Tur} to infer that all positive integers
$k\le x$ have $\omega(k;A,B)$ factors
in the interval
$$
\left[\frac{\log_2 x}{2\phi(B)},~\frac{2\log_2 x}{\phi(B)}\right],
$$
with
$o(x)$ possible exceptions. Because $d\equiv 2 \pmod M$, we have that
$d\not\equiv\pm 1\pmod p$ for all $p&lt;(\log_3 x)/2$. Hence, there exist at
least
$$\mu:=\lfloor(\log_2 x)/(4\log_4 x \phi(B))\rfloor&gt;(\log_2 x)^{1/3}$$
such primes $d$ which
furthermore are congruent to either $1$ or $-1$ modulo $p$ for the {\bf same}
prime $p&gt;(\log_3 x)/2$.

\medskip

We now count how many such $k$'s can there can be. Because of the above
argument, we can write $144k+1=d_1d_2\cdots d_\mu Q&lt; 288x$ (for some
positive integer $Q$), where each $d_j\equiv \pm 1\pmod p$. Thus, the
number of such $k$'s is at most
\begin{eqnarray*}
\sum_{p&gt;(\log_3  x)/2}\frac{288x}{\mu!}\left(\sum_{\substack{d\le x\\
d \, \equiv \, \fmod{A} {B}\\
d\, \equiv \, \pm \fmod{1} {p}}}\frac{1}{d}\right)^\mu
&amp;\ll &amp; x\sum_{p&gt;(\log_3 x)/2}\left(\frac{2e\log_2 x+O(1)}
{\mu\phi(B)(p-1)}\right)^\mu\\
&amp;\ll &amp;  x\sum_{p&gt;(\log_3 x)/2}\left(\frac{O(\log_4 x)}{p}
\right)^{(\log_2 x)^{1/3}}\\
&amp; = &amp; o(x),
\end{eqnarray*}
which completes the proof of Theorem 3.
\end{proof}

\section{Comments and Numerical Results}

Although we proved that $n^2+k$ is powerful for infinitely many $n$'s only for
most integers $k$, we do conjecture that this is actually true for all
integers $k$.
Indeed, fixing a squarefree integer $k$, the probability that $n^2+k$ is
powerful is of the order $\frac{1}{\sqrt{n^2+k}}\approx \frac{1}{n}$ for
large $n$.  This means that we should expect that
\begin{equation}\label{eq:predict}
\#\{n\le x: n^2+k \mbox{ is powerful}\}
\sim\sum_{n\le x}\frac{1}{n}\sim \log x\rightarrow \infty \quad \mbox{as }x\to \infty
\end{equation}
for any squarefree $k$. From our proof it follows indeed that if
$\#\{n:n^2+k \mbox{ is powerful}\}&gt;1$,
then $\#\{n\le x: n^2+k \mbox{ is powerful}\}\gg \log x$.

\medskip

Table 1 (resp. Table 2)  provides, for each integer $1\le k \le 50$, the
smallest known value of $n$ for which $n^2+k$ (resp. $n^2-k$) is a powerful
number without being a perfect square. These values of $n$ were obtained by
finding the minimal solution of $x^2-Dy^2=\pm k$ by considering various
cubefull $D$'s. Those $n&gt;10^9$ may not be the smallest $n$ for which
$n^2\pm k$ is powerful.

\medskip

Given three integers $a,b,c$, one could ask if the polynomial $a n^2+bn+c$ is
powerful for infinitely many integers $n$. Assuming
that  $a n^2+bn+c$ is a powerful number which is not a square, we can then
write $a n^2+bn+c=Dm^2$ with $D&gt;1$ squarefree and $D\divides m$.  We then have
$$
n=\frac{-b\pm \sqrt{4aDm^2+b^2-4ac}}{2a}.
$$
Since $n$ is an integer, there exists an integer $y$ such that
$4aDm^2+b^2-4ac=y^2,$ or, equivalently, $y^2-aD(2m)^2=b^2-4ac$ with $y\equiv
\pm b \pmod{2a}$. But then the existence of one solution implies the existence
of infinitely many. On the other hand, we also get that if there is an
infinity of integers $y$ for which $y^2-b^2+4ac=Dx^2$ where $D&gt;1$ is
squarefree and $2aD \divides x$, then there exist infinitely many $n$'s for which
$a n^2+bn+c$ is a powerful number.
\vskip 15pt


The prediction (\ref{eq:predict}) may at first seem at odd with  the fact that some of the smallest $n$'s obtained in Tables 1 and 2 are huge.  However, the statement ``$n^2+k$ is powerful'' is equivalent to ``$n^2+k=dm^2$ with $d$ squarefree and $d\divides m$''.  Now for any fixed $d$, this last equation is a generalized Pell equation.  While a solution may not exist for some values of $d$, when it does for a particular $d$, it is well known that the smallest solution can be surprisingly large. When solutions exist, it is still possible that none of them will be such that $d \divides m$.  The size of the smallest solution such that $d \divides m$ can also be quite large. There are thus three possible reasons for the huge value of the smallest solution:  a large value of $d$, a large value of the smallest solution to $n^2+k=dm^2$ or a large value of the smallest solution such that $d \divides m$. We investigated this in Table 3, where $n=n_0$ is the smallest solution to $n^2+k=dm^2$ not taking into account the restriction $d \divides m$. It turns out that the surprisingly large values of the smallest $n$ in Table 1 are not due to a very large value of $d$ but rather to a  large value of $n_0$ (see for instance $k=33$), or to the large size of the smallest solution $n_0$ for which $d \divides m$ (see for instance $k=17$).


\centerline{\bf Table 1}
\vskip 10pt

\begin{small}
\noindent
\begin{tabular}{|c|c|c|}\hline
$k$ &amp; $n$ &amp; $n^2+k$ \\ \hline
1 &amp; 682 &amp; $5^3\cdot 61^2$  \\
2 &amp; 5 &amp; $3^3$ \\
3 &amp; 37 &amp; $2^2\cdot 7^3$  \\
4 &amp; 11  &amp; $5^3 $ \\
5 &amp;1879706  &amp; $3^5 \cdot 7^2 \cdot 23^3 \cdot 29^3$ \\
6 &amp; 463 &amp; $5^4\cdot 7^3$\\
7  &amp; 11 &amp; $2^7$ \\
8 &amp; 10 &amp; $2^2\cdot 3^3$ \\
9 &amp; 2046 &amp; $3^2\cdot 5^3\cdot 61^2$ \\
10 &amp; 341881 &amp; $11^3\cdot 9371^2$\\
11 &amp; 31 &amp; $2^2\cdot 3^5$  \\
12 &amp; 74 &amp; $2^4 \cdot 7^3$ \\
13 &amp; 70 &amp; $17^3$ \\
14 &amp; 5519 &amp; $3^3\cdot 5^5\cdot 19^2$ \\
15 &amp; 793 &amp; $2^7\cdot 17^3$ \\
16 &amp; 22  &amp; $2^2\cdot 5^3$ \\
17 &amp; $n_{17}$ &amp; $f_{17}$ \\
18 &amp; 57 &amp; $3^3\cdot 11^2$ \\
19 &amp; 559 &amp; $2^2\cdot 5^7$ \\
20 &amp; 338 &amp; $2^3\cdot 3^3 \cdot 23^2$ \\
21 &amp; $n_{21}$ &amp; $f_{21}$ \\
22 &amp; 503259461 &amp; $47^3\cdot 491^2\cdot 3181^2$ \\
23 &amp; 45 &amp; $2^{11}$ \\
24 &amp; 926 &amp; $2^2\cdot 5^4\cdot 7^3$ \\
25 &amp; 190 &amp; $5^3\cdot 17^2$ \\  \hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}\hline
$k$ &amp; $n$ &amp; $n^2+k$ \\ \hline
26 &amp; 109 &amp; $3^5\cdot 7^2$ \\
27 &amp; 36 &amp; $3^3\cdot 7^2 $ \\
28 &amp; 62 &amp; $2^5\cdot 11^2$ \\
29 &amp; 436 &amp; $3^2\cdot 5^3\cdot 13^2$ \\
30 &amp; 832836278711 &amp; $31^2 \cdot 59^2 \cdot 67^2 \cdot 79^3 \cdot 9679^2$ \\
31 &amp; 63 &amp; $2^5\cdot 5^3$ \\
32 &amp; 88 &amp; $2^5\cdot 3^5$ \\
33 &amp; $n_{33}$  &amp; $f_{33}$ \\
34 &amp; 7037029 &amp; $5^5 \cdot 7^5\cdot 971^2$ \\
35 &amp; 36&amp; $11^3$ \\
36 &amp; 33 &amp; $3^2\cdot 5^3$ \\
37 &amp; $n_{37}$ &amp; $f_{37}$ \\
38 &amp; 5945 &amp; $3^2\cdot 7^3\cdot 107^2$\\
39 &amp; 31&amp; $2^3\cdot 5^3$\\
40 &amp;  52&amp; $2^3\cdot 7^3$ \\
41 &amp; 78 &amp; $5^3\cdot 7^2$ \\
42 &amp; 720025&amp; $13^3 \cdot 31^3 \cdot 89^2$ \\
43 &amp; 22364 &amp; $11^3\cdot 613^2$ \\
44 &amp; 62 &amp; $2^4\cdot 3^5$ \\
45 &amp; 96 &amp; $3^3\cdot 7^3$ \\
46 &amp; 50927 &amp; $5^5\cdot 11^2 \cdot 19^3$ \\
47 &amp; 39 &amp; $2^5\cdot 7^2$ \\
48 &amp; 148 &amp; $2^6\cdot 7^3$ \\
49 &amp; 524 &amp; $5^3\cdot 13^3$ \\
50 &amp; 1325&amp; $3^5\cdot 5^2\cdot 17^2$ \\ \hline
\end{tabular}
\vskip 10pt


\noindent
Here, $n_{17}=1952785824219551870$ with $f_{17}=3^2\cdot 13^3\cdot 367^2
\cdot 7487^2\cdot 5054107013^2$;

\noindent
$n_{21} = 4580728614212333152148$ with $f_{21}=5^2\cdot 31^2\cdot 37^3
\cdot 41611^2\cdot 3155673955493^2$;

\noindent
$n_{33}=2451448196948930$ with $f_{33}=7^2
\cdot 17^3 \cdot 29^3 \cdot 31992951041^2$;

\noindent
$n_{37}=18651116694721032166213875246076$
with $f_{37}=317^3\cdot 10219159057^2\cdot 323370789682598407^2$.
\end{small}
\vskip 20pt


\centerline{\bf Table 2}
\vskip 10pt

\begin{small}
\noindent
\begin{tabular}{|c|c|c|}\hline
$k$ &amp; $n$ &amp; $n^2-k$ \\ \hline
1 &amp; 3 &amp; $2^3$ \\
2 &amp; 11427 &amp; $7^3 \cdot 617^2$ \\
3 &amp; $n_3$ &amp; $f_3$  \\
4 &amp; 6  &amp; $2^5$ \\
5 &amp; 73 &amp; $2^2\cdot 11^3$ \\
6 &amp; 62531004125 &amp; $19^3\cdot 14831^2\cdot 50909^2$ \\
7  &amp; $n_7$ &amp; $f_7$ \\
8 &amp; 20 &amp; $2^3\cdot 7^2$ \\
9 &amp; 15 &amp; $2^3 \cdot 3^3$ \\
10 &amp; $n_{10}$ &amp; $f_{10}$ \\
11 &amp; 56 &amp; $5^5$  \\
12 &amp; 47 &amp; $13^3$ \\
13 &amp; 16 &amp; $3^5$ \\
14 &amp; 33017 &amp; $5^2\cdot 11^3 \cdot 181^2$ \\
15 &amp; 1138 &amp; $109^3$ \\
16 &amp; 68 &amp; $2^9\cdot 3^2$ \\
17 &amp; 23 &amp; $2^9$ \\
18 &amp; 19 &amp; $7^3$ \\
19 &amp; 762488 &amp; $3^2\cdot 5^3 \cdot 127^2 \cdot 179^2$ \\
20 &amp; 146 &amp; $2^4\cdot 11^3$ \\
21 &amp; 1552808&amp; $43^3\cdot 5507^2$ \\
22 &amp; 47 &amp; $3^7$ \\
23 &amp; 6234 &amp; $7^2\cdot 13^3 \cdot 19^2$ \\
24 &amp; 32 &amp; $2^3 \cdot 5^3$ \\
25 &amp; 45 &amp; $2^4\cdot 5^3$ \\  \hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}\hline
$k$ &amp; $n$ &amp; $n^2-k$ \\ \hline
26 &amp; 2537 &amp; $23^5$ \\
27 &amp; 51700 &amp; $13^3\cdot 1103^2 $ \\
28 &amp; 54 &amp; $2^3 \cdot 19^2$ \\
29 &amp; 426 &amp; $7^3 \cdot 23^2$ \\
30 &amp; 83 &amp; $19^3$ \\
31 &amp; 34 &amp; $3^2\cdot 5^3$ \\
32 &amp; 40 &amp; $2^5\cdot 7^2$ \\
33 &amp; 3601 &amp; $2^8 \cdot 37^3$ \\
34 &amp; 948281 &amp; $3^6\cdot 47^3 \cdot 109^2$ \\
35 &amp; 531783519104&amp; $29^3 \cdot 997^2 \cdot 3415409^2$ \\
36 &amp; 42 &amp; $2^6\cdot 3^3$ \\
37 &amp; 73 &amp; $2^2\cdot 3^3 \cdot 7^2$ \\
38 &amp; 16493 &amp; $11^2\cdot 131^3$ \\
39 &amp; $n_{39}$ &amp; $f_{39}$ \\
40 &amp; 632 &amp; $2^3\cdot 3^3 \cdot 43^2$ \\
41 &amp;71 &amp; $ 2^3\cdot 5^4$ \\
42 &amp; 691888331&amp; $13^2\cdot 79^3\cdot 75797^2$ \\
43 &amp; 5016 &amp; $13^2 \cdot 53^3$ \\
44 &amp; 112 &amp; $2^2\cdot 5^5$ \\
45 &amp; 219 &amp; $2^2\cdot 3^2 \cdot  11^3$ \\
46 &amp; 847 &amp; $3^3\cdot 163^2$ \\
47 &amp; 180190 &amp; $53^3 \cdot 467^2$ \\
48 &amp; 94 &amp; $2^2 \cdot 13^3$ \\
49 &amp; 56 &amp; $3^2 \cdot 7^3$ \\
50 &amp; 57135 &amp; $ 5^2\cdot 7^3 \cdot 617^2$ \\ \hline
\end{tabular}
\vskip 10pt

\noindent
Here, $n_3=15503069909027$ with $f_3=13^3\cdot  239^2 \cdot 64866401293^2$;

\noindent
$n_7=85227106679780$ with $f_7=3^3 \cdot 59^3 \cdot 36192438539^2$;

\noindent
$n_{10}=71457130044805582612325294634331$
with $f_{10}=3^3 \cdot 13^3\cdot 43^2\cdot 6823075915494777091540353511^2$;

\noindent
$n_{39}=82716851195974$ with $f_{39}=7^2\cdot 373^3\cdot 29287^2\cdot 56009^2$.

\end{small}


\section{Acknowledgments}

The first author was partially supported by a grant
from NSERC, while the third author was supported by projects SEP-CONACYT
37259-E and 37260-E.

\newpage

\centerline{\bf Table 3}
\vskip 10pt

\begin{center}
\begin{small}
\noindent
\begin{tabular}{|c|c|c|}\hline
$k$ &amp; $d$ &amp; $n_0$ \\ \hline
1 &amp; 5 &amp; 2  \\
2 &amp; 3 &amp; $2$ \\
3 &amp; 7 &amp; 2  \\
4 &amp; 5  &amp; 1 \\
5 &amp; $3\cdot 23 \cdot 29$   &amp; 26258 \\
6 &amp; 7 &amp; 1\\
7  &amp; 2 &amp; 1 \\
8 &amp; 3 &amp; 2 \\
9 &amp; 5 &amp; 6 \\
10 &amp; 11 &amp; 1\\
11 &amp; 3 &amp; 1  \\
12 &amp; 7 &amp; 4 \\
13 &amp; 17 &amp; 2 \\
14 &amp; 15 &amp; 1\\
15 &amp; 34 &amp; 11 \\
16 &amp; 5  &amp; 2 \\
17 &amp; 13 &amp; 10 \\
18 &amp; 3 &amp; 3 \\
19 &amp; 5 &amp; 1\\
20 &amp; 6 &amp; $2$ \\
21 &amp; 37 &amp; 4 \\
22 &amp; 47 &amp; $5$ \\
23 &amp; 2 &amp; 3 \\
24 &amp; 7 &amp; 2 \\
25 &amp; 5 &amp; 10\\  \hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}\hline
$k$ &amp; $d$ &amp; $n_0$ \\ \hline
26 &amp; 3 &amp; 1 \\
27 &amp; 3 &amp; 9 \\
28 &amp; 2 &amp; 2 \\
29 &amp; 5 &amp; 4 \\
30 &amp; 79 &amp; 7 \\
31 &amp; 10 &amp; 3 \\
32 &amp; 6 &amp; 8 \\
33 &amp; $17\cdot 29$  &amp; 1310 \\
34 &amp; 35 &amp; 1 \\
35 &amp; 11&amp; 3 \\
36 &amp; 5 &amp; 3 \\
37 &amp; 317 &amp; 61016 \\
38 &amp; 7 &amp; 5\\
39 &amp; 10 &amp; 1\\
40 &amp;  14 &amp; 4 \\
41 &amp; 5 &amp; 2 \\
42 &amp; $31\cdot 13$ &amp; 19 \\
43 &amp; 11 &amp; 1 \\
44 &amp; 3 &amp; 2\\
45 &amp; 21 &amp; 12 \\
46 &amp; 95 &amp; 7 \\
47 &amp; 2 &amp; 5 \\
48 &amp; 7 &amp; 8 \\
49 &amp; 65 &amp; 4 \\
50 &amp; 3&amp; 5\\ \hline
\end{tabular}
\vskip 10pt
\end{small}
\end{center}
\vskip 30pt

\begin{thebibliography}{9}

\bibitem{kn:mollinwalsh} R. A.\,Mollin and P. G.\,Walsh, Proper differences of nonsquare powerful numbers, {\it C. R. Math. Rep. Acad. Sci. Canada} {\bf 10}
(1988), no.\ 2, 71--76.


\bibitem{MV} H. L.\,Montgomery and R. C.\,Vaughan, The large sieve,
{\it Mathematika\/} {\bf 20} (1973), 119--134.


\bibitem{Pom} C.\,Pomerance, On the distribution of amicable numbers,
{\it J. reine angew. Math.\/}, {\bf 293/294} (1977), 217--222.


\bibitem{rob} J. P.\,Robertson, Solving the generalized Pell 
equation $x^2-Dy^2=N$, \hfill \newline
\href{http://www.hometown.aol.com/jpr2718/pell.pdf}{\tt http://www.hometown.aol.com/jpr2718/pell.pdf}.


\bibitem{Tur} P. Tur{\'a}n,
Uber einige Verallgemeinerungen eines Satzes von Hardy und Ramanujan,
{\it J. London Math. Soc.\/}, {\bf  11} (1936), 125--133.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A05; Secondary 11N25.

\noindent \emph{Keywords: } 
powerful numbers, generalized Pell equation.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A001694}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 27 2010;
revised version received February 22 2011. 
Published in {\it Journal of Integer Sequences}, March 25 2011.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                

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