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P:q = 2:3 then 2P2q = q2 |
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Answer» -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 |
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