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# Proof...

Is the statement true : "if 'p' is a prime then 4 $$p^{2}$$ + 1 is also prime ?"

Note by Vighnesh Raut
3 years, 5 months ago

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Let $$p = 2n + 1$$ as most of the primes are odd, with particular natural number $$n$$. We now have $$16n^{2} + 16n + 5$$ as the number to check. One only needs to choose any multiple of $$5$$ as $$n$$ while maintaining the primality $$p$$. With $$n = 5$$, $$p = 11$$, which is prime, and $$4p^{2} + 1 = 485$$, which isn't a prime.

So, no.

Note: In a glance, I know the statement are false; because if it is true, we won't have a problem in finding the next biggest prime. Which we actually kind of have.

- 3 years, 5 months ago

Firstly your first assumption is wrong Prime number cannot be written as 2n +1 put n=4

You can say 'most' of the prime numbers can be written as $$p = 6n \pm 1$$ Except numbers ending with 5 in the above expression

- 3 years, 5 months ago

First of all, I wasn't nor intended to say that every prime can be written as $$2n + 1$$, with any integer $$n$$. As it is obviously wrong. To clarify, I'm using $$2n + 1$$ as $$p$$ for the sake of one and only one thing, making the problem easier to be solved.

Second, have you read the third sentence in the first paragraph? It says "One only needs to choose any multiple of $$5$$ as $$n$$ while maintaining the primality of $$p$$."

And third, don't write "firstly" if you're only pointing one thing.

- 3 years, 5 months ago

Oh

- 3 years, 5 months ago

$$f(p) = 4p^{2} + 1$$

$$f(p) = 4p^{2} + 4p +1 - 4p$$

$$= (2p + 1)^{2} - 4p$$

odd - even is not always a prime number

therefore statement is false

- 3 years, 5 months ago