For what odd primes $p$ does $p - 2 \mid p + 2$?
For example, $p = 3$.
Are there any others?
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Bill Dubuque
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if you know the answer why do you ask? – Tsemo Aristide Dec 29 '16 at 22:06
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@TsemoAristide, mostly for my own benefit. Most of my questions here stem from considerations related to some problems in number theory. That is why, if I know of an answer to an existing question of mine, I post it right away. – Jose Arnaldo Bebita Dris Dec 29 '16 at 22:08
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1This is only to do feedback about it, from my viewpoint: that could be interesting ask about a possible generalization. I don't know if the generalization in the literature, but maybe is feasible. Thanks. – Dec 29 '16 at 22:31
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1Thank you for your feedback, @user243301. I will keep that in mind. – Jose Arnaldo Bebita Dris Dec 29 '16 at 22:32
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As here in the dupe $,p-2\mid p+2\iff p-2\mid 2+2,$ by mod divisibility reduction. – Bill Dubuque Oct 17 '23 at 05:14
3 Answers
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Notice that $\frac{p+2}{p-2}=1+\frac{4}{p-2}$, which is less than $2$ for $p>6$.
So we must only check $p=3$ and $5$ and only the first works.
Asinomás
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We can look at the problem for any $p\ge2$, regardless of prime or not. Then taking $a:=p-2$ and $b:=p+2$, we can see that $a=b-4$, so $a\mid b \implies a\mid 4$, giving $a=\{1,2,4\}$ as the only possibilities, and thus $p=\{3,4,6\}$ as the corresponding solutions.
Of course only one of these is prime: $p=3$.
Joffan
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Notice that $$\gcd(p - 2, p + 2) = \gcd(4, p + 2) = \gcd(4, p - 2)$$ so that if $p - 2 \mid p + 2$, then $$\gcd(p - 2, p + 2) = p - 2 = \gcd(4, p - 2).$$
Since $p$ is an odd prime then $p \geq 3$ and either $p \equiv 1 \pmod 4$ or $p \equiv 3 \pmod 4$.
Consequently, equality can only occur in $$\gcd(p - 2, p + 2) = p - 2 = \gcd(4, p - 2)$$ when $p = 3$.
QED