Well if we talk about your question if we consider \(N^{2}\)-2, we will get the primes when N=odd number except 1 and if we consider \(N^{2}\)-5 , we will get the primes when N= even number except 2...

You made a good observation that if \(N\) is even, then \( N^2 - 2 \) is even and hence not a prime if \( N > 2 \).

However, this does not imply that if \(N\) is odd, then \( N^2 - 2 \) must be a prime. For example, \( 11 ^ 2 -2 = 119 = 7 \times 17 \). We can show that if \( N \equiv 11 \pmod{14} \) , then \( N^2 - 2 \) is a multiple of 7 and hence not prime.

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TopNewest\(N^{2}-1=(N+1)(N-1)\)So no such primes!! (Except when N-1=1)Similarly for

\(N^{2}-4=(N+2)(N-2)\)No primes!! (Except when N-2=1, Oh but wait! It implies N=3 and \(3^{2}-1=8\) which is not a prime!)What I think is for all perfect squares a, at most a single prime

mayexist!For non perfect squares, I am getting an intuition that infinite primes may exist!

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what about \({ n }^{ 2 }-3\)

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\(4^{2}-3=13\) so it may be a prime number! And I guess maybe ∞ such prime numbers

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Well if we talk about your question if we consider \(N^{2}\)-2, we will get the primes when N=odd number except 1 and if we consider \(N^{2}\)-5 , we will get the primes when N= even number except 2...

In my view this works the best!!!

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You made a good observation that if \(N\) is even, then \( N^2 - 2 \) is even and hence not a prime if \( N > 2 \).

However, this does not imply that if \(N\) is odd, then \( N^2 - 2 \) must be a prime. For example, \( 11 ^ 2 -2 = 119 = 7 \times 17 \). We can show that if \( N \equiv 11 \pmod{14} \) , then \( N^2 - 2 \) is a multiple of 7 and hence not prime.

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7*19=133.

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@Calvin Lin Typoed! 119=7×17

Yeah!Log in to reply

@shivamani patil @Pranjal Jain Thanks! Fixed the typo.

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well infinitely many prime numbers are of the form 6k±1, it will cover all the primes....

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