To prove $\gcd\{p^2-1: p $ is a prime $ \geq 7 \}=24$

Asked Aug 31 '19 at 17:05

Active Aug 31 '19 at 17:09

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Let $p$ be a prime $\geq 7$.

$p^2-1=(p+1)(p-1)$. Both $(p-1)$ and $(p+1)$ are divisible by $2$ and they are two consecutive even numbers. So, either one of the two must be divisible by $4$. So, $8 \mid (p^2-1)$. Again, by Euler's theorem, $p^2 \equiv 1 \pmod3$. Thereby, $24 \mid (p^2-1)$, for all prime $p \geq 7$.

To prove that $24$ is the greatest common divisor:

For $p=7$, $p^2-1=48$. So, the only other possible $\gcd$ is going to be $48$. But for $p=11$, $p^2-1=120$ and $48 \nmid 120$.

Therefore $\gcd\{p^2-1: p $ is a prime $ \geq 7 \}=24$.

Is this correct? Kindly verify.

edited Aug 31 '19 at 17:09

asked Aug 31 '19 at 17:05

Subhasis Biswas

2,600

3

Yes, you are correct. – MafPrivate Aug 31 '19 at 17:08
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https://www.google.com/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3DZMkIiFs35HQ&ved=2ahUKEwiM6Pvsza3kAhUxSN8KHZ4SDJcQwqsBMAB6BAgEEAU&usg=AOvVaw1XoFlHrOW2tdcnMYwLQJj4 – Aug 31 '19 at 17:14
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numberphile has a video on this problem – Donlans Donlans Aug 31 '19 at 17:24
Whoah... checked the video. :) – Subhasis Biswas Aug 31 '19 at 17:27
Umm, doesn't a greatest common factor require two numbers? If one is $p^2-1$, what's the other? – Rhys Hughes Aug 31 '19 at 17:33
@RhysHughes, we are taking the gcd of all of the primes of the form $p^2-1$, $p$ being the set of all primes $\geq 7$ – Subhasis Biswas Aug 31 '19 at 17:34
gcd of primes is 1. You can't even state your own question correctly. – Aug 31 '19 at 17:55
of all the numbers of the form $p^2-1$. If $p$ is a prime larger than $2$, $p^2-1$ is never a prime. Ignore the typo.

Subhasis Biswas

Aug 31 '19 at 18:11

To prove $\gcd\{p^2-1: p $ is a prime $ \geq 7 \}=24$

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