Primes....

I have been recently thinking about the squares of prime numbers "p" and found that the number of the form p^2 - 1 is divisible by 12 except the primes 2 and 3. Is my statement correct ? So , if it is correct can anyone prove how are they divisible by 12?

#RealNumbers #NumberThoery

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

p^2-1 can be written as (p-1)(p+1).2 is the only even prime number.Hence,p+1,p-1 are even.Thus,(p+1)(p-1)is divisible by 4.Also,if we take any three consecutive natural numbers,one of them is divisible by three.Here,if we take (p-1),p,(p+1),then,p is not divisible by 3 as it is prime,thus,either (p+1) or (p-1) should be divisible by 3.Hence,p^2-1 is always divisible by 12 for prime numbers 5 and above.

Akshay Bodhare - 6 years, 5 months ago

Thanks for your proof

Sriram Venkatesan - 6 years, 5 months ago

It is simple.No problem.

Akshay Bodhare - 6 years, 5 months ago

mADE ME THINK...

vishwathiga jayasankar - 6 years, 5 months ago

$p^2 -1 =(p-1)(p+1)$ . Now all primes satisfy: $(p\equiv1\mod6)\ or \ (p\equiv-1\mod6)$ , (as the other options can be factorized) which means that in both cases we have 6|(p-1)(p+1). Also, both of ${p}$ -1 and ${p}$ +1 are divisible by two as ${p}$ is odd ( for ${p}$ > 2). $\therefore$ 24| $p^{2}$ -1 $\Rightarrow$ 12| $p^{2}$ -1

Curtis Clement - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$