Question

P:q = 2:3 then 2P2q =
q2
​

Answer 1

Step-by-step explanation:

First, we square the first equation. We have: p2q2=r2+2r+1.

If we plug r2+1 we would have: p2q2=2(p2+q2+r). Since p and q are prime numbers and the RHS is even, at least one of them should be 2.

Without loss of generality we take p=2. We have: 4q2=2(4+q2+r) so q2=r+4. If we plug this into the second equation, we have: 2(4+r+4)=r2+1 so r2−2r+15=0.

The solutions are r=−3,5 but −3 is not a prime number so r=5. Plugging this into the first equation, we get 2q=6 so q=3.

Thus, the only triplets would be: (p,q,r)=(2,3,5),(3,2,5)

P.S.: As Singhal suggested, Technically −3 is a prime. But in here it does not lead to any solutions