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Problem. Let $(a_n)_{n \geq 0}\subseteq \mathbb{Z}$, defined by: $a_n=(n^2+3)(n^2-13)(n^2+39), \forall n \in \mathbb{N}.$ Show that for any natural number $m \geq 2$, there exists $n\in \mathbb{N}$, such that $m|a_n$.

This problem was proposed a long time ago for an JBMO selection, but unfortunately I don't have neither the exact source nor the official solution. Therefore, I have come up with my own.

Please check these arguments (maybe with all the details, if possible), and any suggestions, improvements, or even shorter, ellegant solutions are more than welcome. Thank you for your time and effort!

Before solving, first some additional considerations about quadratic residues:

Proposition.
a) Let $p$ be a prime natural number and $a\in \mathbb{Z}, p\nmid a$. Then $a$ is quadratic residue modulo $p$ iff $a$ is quadratic residue modulo $p^k$, for any natural number $k\geq 1$.
b) Let $a$ be an odd integer, and $k\geq 3$ be a fixed natural number. Then $a$ is quadratic residue modulo $2^k$ iff $a\equiv 1$ (mod $8$).

Proof:
$a)$ As the converse is obvious, it remains to show the direct implication, by induction over $k\geq 1$. Let $n_1\in \mathbb{N}$, for which $p^1|n_1^2-a$, this is the verification step. Suppose that for a certain $k\geq 1$ we have constructed $n_k\in \mathbb{N}$, for which $p^k|n_k^2-a$, and we want to find a $n_{k+1 }\in \mathbb{N}$, for which $p^{k+1}|n_{k+1}^2-a$. For this, we will look for $\lambda \in \mathbb{N}$, with $n_{k+1}=n_{k}+\lambda p^k$. So:
$$p^{k+1}|n_{k+1}^2-a \Leftrightarrow p^{k+1}|n_k^2-a+\lambda^2p^{2k}+2\lambda n_kp^k \Leftrightarrow p|x_k+2\lambda n_k, \tag{1}$$ where $x_k:=(n_k^2-a)/p^k \in \mathbb{Z}$. If somehow $p|n_k$, from $p|p^k|n_k^2-a$ would result $p|a$, contradiction with the hypothesis! So $p\nmid n_k, p\nmid 2 \Rightarrow p\nmid 2n_k$. Thus:
$$(1) \Leftrightarrow (2n_k)\lambda \equiv -x_k (\text{mod p}) \Leftrightarrow \lambda \equiv -x_k(2n_k)^{-1} (\text{mod p}). \tag{2}$$ We choose $\lambda\in\mathbb{N}$, which verifies the condition from $(2)$, and the induction is complete.

$b)$ ($\Rightarrow$) If $a$ is quadratic residue modulo $n$, then it is quadratic residue modulo any divisor of $n$. Here, $a$ will become a quadratic residue modulo $2^3=8$. As $a$ is odd, and the quadratic residues modulo $8$ are $0,1,4$, it follows that $a\equiv 1$ (mod $8$).
($\Leftarrow$) Let $a\equiv 1$ (mod $8$), then, as above, $a$ is quadratic residue modulo $8$. We will show (similar to point $a)$ by induction after $k\geq 3$ that $a$ is quadratic residue modulo $2^k$, for any such $k$. The verification step is already done, and $n_1\in \mathbb{N}$ appears, for which $2^3|n_1^2-a$. Suppose that for a certain $k\geq 3$ we have constructed $n_k\in \mathbb{N}$, for which $2^k|n_k^2-a$, and we want to find a $n_{k+1} \in \mathbb{N}$, for which $2^{k+1}|n_{k+1}^2-a$. For this, we will look for $\lambda \in \mathbb{N}$, with $n_{k+1}=n_{k}+\lambda 2^{k-1}$ ($2^k$ is not enough, since $2$ is non-invertible modulo $p=2$). So:
$$2^{k+1}|n_{k+1}^2-a \Leftrightarrow 2^{k+1}|n_k^2-a+\lambda^2p^{2k-2}+\lambda n_k2^k \Leftrightarrow 2|x_k+\lambda n_k, \tag{3}$$ where $x_k:=(n_k^2-a)/2^k \in \mathbb{Z}$. If somehow $2|n_k$, from $2|2^k|n_k^2-a$ would result $2|a$, contradiction with the fact that $a$ is odd! So $2\nmid n_k$, and then: $$(3) \Leftrightarrow n_k\lambda \equiv -x_k (\text{mod 2}) \Leftrightarrow \lambda \equiv -x_k (\text{mod 2}). \tag{4}$$ We choose $\lambda\in \{0,1\}$, which verifies the condition from $(4)$, and the induction is complete. $\Box$

Paul Rebenciuc
  • 931

1 Answers1

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Problem. Let $(a_n)_{n \geq 0}\subseteq \mathbb{Z}$, defined by: $a_n=(n^2+3)(n^2-13)(n^2+39), \forall n \in \mathbb{N}.$ Show that for any natural number $m \geq 2$, there exists $n\in \mathbb{N}$, such that $m|a_n$.

We will show the following intermediate result: if $m_1, m_2 \geq 2$ are two coprime natural numbers, with $m_1|a_i, m_2|a_j$, for some $i,j \in \mathbb{N }$, then $m_1m_2|a_k$ for a certain $k$ natural number.

We can assume, without loss of generality, that $i\geq j$. It is clear that $m_1|a_{i+xm_1}, m_2|a_{j+ym_2}$, for any $x,y \in \mathbb{N}$. As $(m_1, m_2)=1$, there exists $r\in \mathbb{N}$, for which $m_1r \equiv 1$ (mod $m_2$) (modular inverse). It is enough to show that the equation $i+xm_1=j+ym_2$ has solutions in $x,y$, in which case we will take $k:=i+xm_1$. Indeed, the equation becomes:
$i-j+xm_1 \equiv 0$ (mod $m_2$) $\Leftrightarrow x\equiv (j-i)r $ (mod $m_2$). We choose an $x\in \mathbb{N}$ with this property, then $y=\frac{i-j+xm_1}{m_2}\in \mathbb{N}$, hence the conclusion.

Due to the above and prime factorization, it remains to prove for $m=p^k$, i.e. for powers of prime numbers. To solve the problem, it is enough to show that for such $m$, at least one of $-3,13,-39$, is a quadratic residue modulo $m$ (in which case there is a natural $n$, with $ p^k$ divides $n^2+3$ or $n^2-13$ or $n^2+39$, so $p^k|a_n$).

We show that the statement holds for $k=1$, first. We have that: $a_1=-1920, a_6=67275$, so $2|a_1, 3|a_1, 13|a_6$. Let from now $p\neq 2,3,13$, and assume by absurdity that $p\nmid a_n, \forall n\in \mathbb{N}$. This means that $p \nmid n^2+3, n^2-13, n^2+39, \forall n\in \mathbb{N}$, that is: $$\left(\frac{-3}{p}\right)=\left(\frac{13}{p}\right)=\left(\frac{-39}{p}\right)=- 1 \Rightarrow \left(\frac{-39}{p}\right)=\left(\frac{-3}{p}\right)\left(\frac{13}{p}\right)=(-1)(-1)=1,$$ contradiction. So, $p|a_n$, for a certain natural number $n$.

Thus, we deduce that, for any prime number $p\neq 2,3,13$, at least one of $-3,13,-39$, is quadratic residue modulo $p$. Also, $\left(\frac{13}{3}\right)=\left(\frac{-3}{13}\right)=1.$ Therefore, for any prime number $p\neq 2$, at least one of $-3,13,-39$, is quadratic residue modulo $p$. According to point $a)$ of the Proposition, at least one of $-3,13,-39$, is quadratic residue modulo $p^k$, for any $k\geq 1$ natural number, hence the conclusion of the problem.

We are left with the case $p=2$. Of course, it suffices to prove for $2^k, k\geq 3$. We have that $-39\equiv 1$ (mod $8$), and according to point $b)$ of the Proposition, $-39$ becomes a quadratic residue modulo $2^k$, for any $k\geq 3$, hence the conclusion.

Generalization. We consider the sequence $(x_n)_{n \geq 0}$ of integers, defined by: $x_n=(n^2-a)(n^2-b)(n^2-ab), \forall n \in \mathbb{N}.$ For what unassociated odd primes $a,b\in \mathbb{Z}$, it is true (with the same proof as above) that: for any $m \geq 2$ natural number, there exists $n\in \mathbb{N}$ such that $m|x_n$?

Solution: The previous statement holds for $a=-3, b=13$. The intermediate result concerning the reduction to powers of prime numbers was done regardless of $a,b$ (all that mattered is the polynomial form of $x_n$ as a function of $n$). Then, it sufficed to show that for each $m=p^k$, one of $a,b$ or $ab$ is quadratic remainder modulo $m$.

For $k=1$, $p\neq 2,a,b$, the argument with the Legendre symbol is valid, $a|x_{|a|}, b|x_{|b|}, 2|x_1$. So we solved for $p^k$, with $p\neq 2,a,b$.

The argument for $2^k$ works if $a,b$ or $ab$ is congruent to $1$, modulo $8$. Also, for $a^k$, we need $b$ to be a quadratic residue modulo $|a|$. Similarly, for $b^k$, $a$ has to be a quadratic residue modulo $|b|$.

In conclusion, the proposed solution is essentially the same, for non-associated odd primes $a,b\in \mathbb{Z}$, which simultaneously check the conditions: \begin{equation} \begin{cases} a \equiv 1 (\text{mod 8}) \lor b \equiv 1 (\text{mod 8}) \lor ab \equiv 1 (\text{mod 8})\\ \left(\frac{a}{|b|}\right)=\left(\frac{b}{|a|}\right)=1. \end{cases} \end{equation}

Paul Rebenciuc
  • 931