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# Help!!!

Pls help me solving this problem.

Note by Shivamani Patil
2 years ago

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$$N^{2}-1=(N+1)(N-1)$$ 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 may exist!

For non perfect squares, I am getting an intuition that infinite primes may exist! · 1 year, 12 months ago

what about $${ n }^{ 2 }-3$$ · 1 year, 11 months ago

$$4^{2}-3=13$$ so it may be a prime number! And I guess maybe ∞ such prime numbers · 1 year, 11 months ago

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!!! · 2 years ago

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. Staff · 1 year, 12 months ago

7*19=133. · 1 year, 11 months ago

Yeah! @Calvin Lin Typoed! 119=7×17 · 1 year, 11 months ago

@shivamani patil @Pranjal Jain Thanks! Fixed the typo. Staff · 1 year, 11 months ago