When is the integer between twin primes a perfect square? Explain why there exists an integer $n$ such that $4n^2 − 1 = p$ Suppose that $$ and $ + 2$ are both prime numbers.
a) Is the integer between $$ and $ + 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 $$
and $ + 2$ is a perfect square.
Also $(2n)^2=4n^2, (2n+1)^2=4n^2+4n+1$
Explain why there exists an integer $n$ such that $4^2 − 1 = .$
c) By considering $(2 − 1)(2 + 1),$ find the only possible value of $.$
Part (a) seems straight forward, although I imagine there's a more formal explanation required?
Any tips on parts (b) and (c), any obvious deductions/observations i need to consider?
 A: (b) is straightforward: $p + 1$ can be written as $x ^ 2$, so $p \ne \pm 2$ otherwise that perfect square couldn't be written as $(2n)^2$. As you said $p + 1 = 4n^2$, subtract $1$ from bot sides: $p = 4n^2 - 1$.
(c) As @lulu said $2n - 1 = 1 \Rightarrow n = 1 \Rightarrow 4n^2 - 1 = 3$
A: (a) If $p=2$ then $p+2=4$ is not prime, i.e. we can be sure that the two primes $p$ and $p+2$ are odd.
(b) If $p+1=n^2$ with positive integer $n$, then $p=n^2-1=(n+1)(n-1)$ is a factorisation of $p$. Conclude that $n-1=1$, i.e., $p=3$
It turns out that in the only possible case, the number between 3 an 5 is also of the form $4n^2$.
A: Given$\quad A^2+B^2=C^2\quad $ the most common means of generating Pythagorean triples is Euclid's formula
$A=m^2-k^2,\quad B=2mk, \quad C=m^2+k^2.\quad$ We can restrict this formula to generating only primitive triples and odd square multiples of primitives if we let
$\quad m=(2n-1+k),\quad $ i.e.
$A=(2n-1+k)^2-k^2,\quad
 B=2(2n-1+k)k, \quad 
C=(2n-1+k)^2+k^2.$
Which expands to
\begin{align*}
A=&(2n-1)^2+&2(2n-1)k&\\
B=&         &2(2n-1)k& +2k^2\\
C=&(2n-1)^2+&2(2n-1)k&+2k^2
\end{align*}
and, if we let $\space k=1,\space$ the formula collapses to
$$A=4n^2-1\quad B=4n\quad C=4n^2+1$$
Which generates all triples where $\space C-A=2\space$, et seq
$$(3,4,5)\quad (15,8,17)\quad (35,12,37)
\quad (63,16,65)\quad (99,20,101)\quad \cdots
$$
note that, for all of these, $C-A=2$
and
$\dfrac{C+A}{2}$ is a perfect square
If we factor A we get
$$
A=(2n-1)^2+2(2n-1)k\quad = (2 n - 1) (2 k + 2 n - 1)
$$
and we can see that A is always composite unless $n=1$, leaving the only solution where $P$ and $P+2$ can both be prime is  $(3,4,5)$.
