Subfield: Number Theory


Question: Let $S$ be the set of all positive integers $n$ such that $n^{4}$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$. Prove that there are infinitely many elements of $S$ of each of the forms $7 m, 7 m+1,7 m+2,7 m+5,7 m+6$ and no elements of $S$ of the form $7 m+3$ or $7 m+4$, where $m$ is an integer.

Solution:
The conclusion is a consequence of the lemma below which actually provides a recursive description of $S$. The proof of the lemma is at the end of the solution.

Lemma. The fourth power of a positive integer $n$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$ if and only if at least one of the numbers $2 n^{2}+1$ and $12 n^{2}+9$ is a perfect square.

Consequently, a positive integer $n$ is a member of $S$ if and only if $m^{2}-$ $2 n^{2}=1$ or $m^{2}-12 n^{2}=9$ for some positive integer $m$.

The former is a Pell equation whose solutions are $\left(m_{1}, n_{1}\right)=(3,2)$ and

$$
\left(m_{k+1}, n_{k+1}\right)=\left(3 m_{k}+4 n_{k}, 2 m_{k}+3 n_{k}\right), \quad k=1,2,3, \ldots
$$

In what follows, all congruences are modulo 7. Iteration shows that $\left(m_{k+3}, n_{k+3}\right) \equiv\left(m_{k}, n_{k}\right)$. Since $\left(m_{1}, n_{1}\right) \equiv(3,2),\left(m_{2}, n_{2}\right) \equiv(3,-2)$, and



$\left(m_{3}, n_{3}\right) \equiv(1,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and \pm 2 modulo 7 .

The other equation is easily transformed into a Pell equation, $m^{\prime 2}-$ $12 n^{\prime 2}=1$, by noticing that $m$ and $n$ are both divisible by 3 , say $m=3 m^{\prime}$ and $n=3 n^{\prime}$. In this case, the solutions are $\left(m_{1}, n_{1}\right)=(21,6)$ and

$$
\left(m_{k+1}, n_{k+1}\right)=\left(7 m_{k}+24 n_{k}, 2 m_{k}+7 n_{k}\right), \quad k=1,2,3, \ldots
$$

This time iteration shows that $\left(m_{k+4}, n_{k+4}\right) \equiv\left(m_{k}, n_{k}\right)$. Since $\left(m_{1}, n_{1}\right) \equiv$ $(0,-1),\left(m_{2}, n_{2}\right) \equiv(-3,0),\left(m_{3}, n_{3}\right) \equiv(0,1)$, and $\left(m_{4}, n_{4}\right) \equiv(3,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and $\pm 1$ modulo 7 .

Finally, since the $n_{k}$ from the two sets of formulae exhaust $S$, by the preceding no integer in the residue classes $\pm 3$ modulo 7 is a member of $S$.

We now turn to the proof of the lemma. Let $n$ be a member of $S$, and let $d=n^{2}+m$ be a divisor of $n^{4}$ in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$, so $1 \leq m \leq 2 n$. Consideration of the square of $n^{2}=d-m$ shows $m^{2}$ divisible by $d$, so $m^{2} / d$ is a positive integer. Since $n^{2}<d<(n+1)^{2}$, it follows that $d$ is not a square; in particular, $m^{2} / d \neq 1$, so $m^{2} / d \geq 2$. On the other hand, $1 \leq m \leq 2 n$, so $m^{2} / d=m^{2} /\left(n^{2}+m\right) \leq 4 n^{2} /\left(n^{2}+1\right)<4$. Consequently, $m^{2} / d=2$ or $m^{2} / d=3$; that is, $m^{2} /\left(n^{2}+m\right)=2$ or $m^{2} /\left(n^{2}+m\right)=3$. In the former case, $2 n^{2}+1=(m-1)^{2}$, and in the latter, $12 n^{2}+9=(2 m-3)^{2}$.

Conversely, if $2 n^{2}+1=m^{2}$ for some positive integer $m$, then $1<m^{2}<$ $4 n^{2}$, so $1<m<2 n$, and $n^{4}=\left(n^{2}+m+1\right)\left(n^{2}-m+1\right)$, so the first factor is the desired divisor.

Similarly, if $12 n^{2}+9=m^{2}$ for some positive integer $m$, then $m$ is odd, $n \geq 6$, and $n^{4}=\left(n^{2}+m / 2+3 / 2\right)\left(n^{2}-m / 2+3 / 2\right)$, and again the first factor is the desired divisor.