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These are two problems which I have been trying to solve.

The arithmetic mean of twin primes 5 and 7 is 6 which is a triangular number. Do there exist any other such twin primes? If they exist find a pair otherwise prove that there do not exist any other such twin primes.

Let the smaller prime be $p$. The larger one is $p+2$. Their mean is $p+1$. Triangular numbers are of the form $\frac{n(n+1)}{2}$. So,

$$\frac{p+p+2}{2} = \frac{n(n+1)}{2}$$

How do I proceed further?

The arithmetic mean of twin primes 3 and 5 is 4 which is a perfect square. Do there exist any other such twin primes? If they exist find a pair otherwise prove that there do not exist any other such twin primes.

I have made no progress in this one.

Thanks.

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2 Answers 2

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On the first one:

$$p+1 = \frac{n(n+1)}{2}$$ $$p = \frac{n(n+1)}{2} - 1$$ $$p = \frac{(n^2 + n - 2)}{2}$$ $$p = \frac{(n-1)(n+2)}{2}$$

Since $p$ is prime, $(n-1)/2 = 1$, so $n=3$ and $p+1 = n(n-1)/2 = 3*4/2 = 6$; or $n-1 = 1$ so $p+1 = 3$ (which, when we go back and check, is not a solution as $p+2 = 4$ and we're looking at twin primes). This proves $p+1 = 6$ is the only average of twin primes that is a triangular number.

On the second one:

$$p+1 = n^2$$ $$p = n^2 - 1$$ $$p = (n-1)(n+1)$$

Since $p$ is prime, $n-1 = 1$. This yields $p+1 = n^2 = 2^2 = 4$. This proves $p+1 = 4$ is the only average of twin primes that is a perfect square.

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For the second problem:

what form does a perfect square take? ($n^2$)

the prime number less than it is then: $n^2-1$, which has the factorization

$n^2-1=(n-1)(n+1)$

This number ($n^2-1$) is prime iff $(n-1)$ and $(n+1)$ are one and a prime, which is the case for $n = 2$ (the example you provide), but are certainly not for any other $n$

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    $\begingroup$ I got the idea but don't agree with 'This number ($n^2-1$) is prime iff $(n-1)$ and $(n+1)$ are both prime'. The multiple of two prime isn't a prime. It should say that at least one of them is 1which happens in case $n=2$. $\endgroup$ Mar 24, 2016 at 16:20
  • $\begingroup$ @Dhruv right - I didn't mean to imply that e.g. 29x31 was prime just because they were each prime $\endgroup$
    – costrom
    Mar 24, 2016 at 16:33

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