Number of triplets of primes such that `pq =r+1` and `2(p^2 + q^2) = r^2 + 1` is

Find all triples (p, 9,r) of primes such that pq = r + 1 and 2(p^2+ q^2) =r^2 + 1

The number of all possible triplets (p,q,r) such that p+q cos2 theta+r sin^(2)theta=0 for all theta is

Factorise the following : (p^2 - 2pq+ q^2) - r^2 .

If p, q, r, s are in G.P, show that (p^2 + q^2 + r^2) (q^2 + r^2 + s^2) = (pq + qr + rs)^2

lf r, s, t are prime numbers and p, q are the positive integers such that their LCM of p,q is r^2 t^4 s^2, then the numbers of ordered pair of (p, q) is

lf r, s, t are prime numbers and p, q are the positive integers such that their LCM of p,q is r^2 t^4 s^2, then the numbers of ordered pair of (p, q) is

If p,q and r are prime numbers such that r=q+2 and q=p+2 , then the number of triples of the form (p,q,r) is

For all positive values of p,q,r and s ((p^(2) + p +1)(q^(2) + q+ 1)(r^(2) + r+1)(s^(2) + s + 1))/(pqrs) cannot be less than