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This comes from some university's admission test. Here's a translation of an excerpt of a problem:

A positive integer $k$ is a cool number if for every natural number $n$ coprime with $k$, the quantity $n^2 - 1$ is divisible by $k$.

The original problem asked for the biggest cool number below a certain threshold, but I was interested in finding all of them. I managed to prove that $k \in \lbrace 1, 2, 3, 4, 6, 8, 12, 24 \rbrace$ satisfies the contidion, but, to my surprise, I couldn't find more! Crunching some numbers on a computer revealed that these are the only cool numbers less than $10^9$. This has led me to the following conjecture:

$$\lbrace 1, 2, 3, 4, 6, 8, 12, 24 \rbrace \mathrm{\ are \ the \ only \ } \mathit{cool} \mathrm{\ \, numbers.}$$

Beautiful, isn't it? I've attempted to prove this by way of contradiction. My idea was, assume there is a cool $k > 24$, get an upper bound $M$ on it, show one-by-one that no natural number $n$ s.t. $24 < n < M$ satisfies the condition, end the proof. Unfortunately, I can't find a way to attack this problem. I've noticed a neat little fact about the conjectured cool numbers, they're the divisors of 24, but this may be a coincidence and I don't see how I could use this fact.

Bill Dubuque
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1 Answers1

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Claim: If $p$ divides $k$ with $p>3$ then $k$ is not cool.

Proof: Take $n$ coprime with $k$ such that $n$ is congruent to $2\bmod p$ (possible by the chinese remainder theorem) and notice $p\nmid n^2-1$.

Claim: If $9$ divides $k$ then $k$ is not cool.

Proof: Take $n$ coprime with $k$ such that $n$ is congruent to $2\bmod 9$ (possible by the chinese remainder theorem) and notice $9\nmid n^2-1$ so $k\nmid n^2-1$

Claim: If $16$ divides $k$ then $k$ is not cool.

Proof: Take $n$ coprime with $k$ such that $n$ is congruent to $3\bmod 16$ (possible by the chinese remainder theorem) and notice $16\nmid n^2-1$ so $k\nmid n^2-1$

Finally one can check that $2,4,8$ and $3$ work by inspection, so by the chinese remainder all of the following work,

$\{1,3,2,6,4,12,8,24\}$

Asinomás
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