Twin primes and squares

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

If q and p are 3 and 5 respectively then we know that my conjecture holds true [ ( 3 X 5 ) + 1 ] = 16 =4 X 4.

Now consider the case when q is not equal to 3. Now, we can check the remainders of q and p when divided by 3. If q = 0 mod 3 then contradiction.( q is prime and not equal to 3) If q = 1 mod 3 then p = 0 mod 3 ( which is again a contradiction ( p is prime)) If q = 2 mod 3 then p = 1 mod 3

This means that q = 3m-1 and p = 3m+1 so pq = (9m^2) - 1 and pq + 1 = 9m^2.

So a (somewhat) stronger conjecture would be "For all such primes not equal to 3, their (product + 1) is always a square divisible by 9".

$\begin{aligned} p \blue q + 1 & = p \blue{(p-2)} + 1 & \small \blue{\text{Since }p-q=2 \implies q = p-2} \\ & = p^2 - 2p + 1 \\ & = (p-1)^2 & \small \blue{\boxed{\text{A perfect square.}}} \end{aligned}$