How many primes in the first 6000 primes have a particular property How many prime numbers are there in the first $6000$ prime numbers that are the quotients of other prime numbers in the following way  $(P_1^2-1)/(P_2^2-1)=P_3$
where  $P_1$  , $P_2$  and $P_3$   are  different prime numbers.
 A: Some observations: With the restriction that $p_1$ and $p_2$ be in the first $6000$ primes, Charles is of course correct. Here it it in one line of Mathematica:
Union[Select[
        Flatten[
           Table[(Prime[j]^2 - 1)/(Prime[i]^2 - 1), {j, 1, 6000}, {i, 1, j}]], 
        PrimeQ]]

Without this restriction, it is hard to prove that there is any prime NOT of this form. For any prime $p$, there are infinitely many solutions to the equation $(x^2-1)/(y^2-1)=p$ without the requirement that $p$ is prime. Namely, the given equation is equivalent to $x^2 - p y^2 = 1-p$ (except that $(1,1)$ is a solution to the latter and not the original equation.) Let $(u,v)$ solve Pell's equation $u^2-p v^2 = 1$. Then taking $x+y \sqrt{p}= (1+\sqrt{p})(u+v \sqrt{p})^n$, we have $u^2-p v^2 = p-1$. There might also be other solutions, depending on the prime factorization of $p-1$.
For $p=13$, I believe that all solutions to $x^2-13 y^2 = -12$ are of the form $x+\sqrt{13} y = ((11+3 \sqrt{13})/2)^n (1+\sqrt{13})$. For the first $30$ values of $n$, which gets me up to $30$ digits numbers, none of the pairs $(x,y)$ are $(\mathrm{prime}, \mathrm{prime})$. (EDIT: PeterKošinár above points out that I should also check negative $n$, which I haven't gotten around to doing systematically.) However, I see no obstacle to them being so.
I can't even figure out whether or not I expect there to be infinitely many such primes.
