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How many different positive integers n are there for which n and n^2+3 are both prime numbers?

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Answer: 1

That value being n = 2

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Step-by-step explanation:

The only even prime number is n = 2. This leads to n^2+3 = 2^2+3 = 7 which is also prime. So far we found one such value of n that fits the criteria.

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If n is odd, then n^2 is also odd.

The quick proof goes like this: let n = 2k+1 for some integer k. This guarantees n to be odd. Then n^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1 = 2*(some integer)+1 = some odd integer.

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If n is prime and n isn't equal to 2, then n must be odd.

If n was even and n > 2, then it would be a multiple of 2 by definition, and would rule out it being prime. Example: n = 16 is not prime since 2 is a factor.

So n must be odd if we want to find more numbers to fit the given criteria. That makes n^2 being odd as well from the previous section's proof.

But then we use the idea that:

odd + odd = even

I'll leave the proof of this to the reader.

The sum of any two odd numbers gives an even number. An example would be 3+5 = 8.

  • n is prime and not 2 ---> n is odd --> n^2 is odd ---> n^2+3 = odd+odd = even
  • n^2+3 being even means n^2+3 cannot possibly be prime.

There are no odd values of n that are prime and n^2+3 is prime.

We therefore can conclude there's only one value of n that is prime and n^2+3 is also prime. That value being n = 2.

User Egy Mohammad Erdin
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