1

Find the number of natural numbers $n$ for which $n^2-9$ divides $6n^2$.

Seems to me that $n=6$ is the only solution. I wrote a simple code to verify it. Can anyone help me justify why this is the case.

Bijesh K.S
  • 2,604
  • 1
    $\frac{6k}{k-9}=6+\frac{54}{k-9}$ which leaves you with not too many options. – Michal Adamaszek Jan 11 '24 at 13:52
  • You can see after the first few values it ranges between $7$ and $6$, and slowly converges to $6$ but using $$\lim_{x\rightarrow \infty}\frac{6x^2}{(x+3)(x-3)} = 6$$ see that it can't be an integer for any other value, hence nothing other than $n=6$ – Mako Jan 11 '24 at 14:19
  • 1
    Note that $n=0$ also satisfies the divisbility condition. Whether that counts towards the solutions depends on your convention of natural numbers. – anankElpis Jan 11 '24 at 14:37

1 Answers1

1

We have $$n^2 - 9 \mid 6n^2 \Longleftrightarrow n^2 - 9 \mid 6n^2 + k(n^2 - 9) $$ for any $k \in \mathbb{Z}.$ Setting $k = -6$ gives $$n^2 - 9 \mid 54$$ Since $n \in \mathbb{N}$, this also means $n^2 - 9 \le 54$ and thus $$1 \le n \le 7$$ Checking all possible values, we only get $n = 6$.

SchellerSchatten
  • 160