If p, q, r are prime numbers such that p^2-q^2=r. Find all the possible ordered pairs.

Dear student,
r=(p-q)(p+q)
so (p-q) and (p+q) are factors of r
But r is prime so it has only 2 factors ie 1 and r
As, r>1, So,
p-q=1. ————1
p+q=r. ————2
because p+q>p-q
Adding equation 1 and 2
2p=r+1
=>p=(r+1)/2
Putting p=(r+1)/2 in equation 2
q=r-[(r+1)/2)
=>q=(r-1)/2
So there are infinitely many solution as r is a prime number.
As, two is the only even prime no.
Thus, p and q will divisible by 2 for all for r except 2
Regards