Suppose that $𝑝$ and $𝑝 + 2$ are both prime numbers.
a) Is the integer between $𝑝$ and $𝑝 + 2$ odd or even? Explain your answer.
All prime numbers are odd except for $\pm 2.$
So the integer between the primes $p$ and $p + 2$ i.e. $p + 1,$ is always even for $p \neq \pm 2;$
and odd for $p = \pm 2.$
b) Assume additionally that the integer between the primes $𝑝$
and $𝑝 + 2$ is a perfect square.
Also $(2n)^2=4n^2, (2n+1)^2=4n^2+4n+1$
Explain why there exists an integer $n$ such that $4𝑛^2 − 1 = 𝑝.$
c) By considering $(2𝑛 − 1)(2𝑛 + 1),$ find the only possible value of $𝑝.$
Part (a) seems straight forward, although I imagine there's a more formal explanation required?
Any tips on parts (b) and (c), any obvious deductions/observations i need to consider?
Best Answer
(a) If $p=2$ then $p+2=4$ is not prime, i.e. we can be sure that the two primes $p$ and $p+2$ are odd.
(b) If $p+1=n^2$ with positive integer $n$, then $p=n^2-1=(n+1)(n-1)$ is a factorisation of $p$. Conclude that $n-1=1$, i.e., $p=3$
It turns out that in the only possible case, the number between 3 an 5 is also of the form $4n^2$.