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Product + Square = Difference of Squares (Posted on 2011-07-03) Difficulty: 3 of 5
Three positive integers P, Q and R, with P < Q < R, are in arithmetic sequence satisfying :
N*P*Q*R + Q2 = R2 - P2, where N is a positive integer.

Determine all possible quadruplet(s) (P, Q, R, N) that satisfy the above equation, and prove that no other quadruplet satisfies the given conditions.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Another approach (Spoiler) Comment 2 of 2 |
NPQR + Q2 = R2 - P2  gives          NP = (R2 - P2 - Q2)/(QR)

Writing  R = 2Q - P  and simplifying then gives      NP = (3Q - 4P)/(2Q - P)

which can be rearranged to                                 NP = 3/2  -  5P/(4Q - 2P)
 
Since 4Q > 2P, it follows that NP < 3/2 and, since N and P are both positive integers, we have N = P = 1 as the only possibility.

Substituting then gives P, Q, R, N  to be 1, 3, 5, 1 respectively.



  Posted by Harry on 2011-07-05 19:15:11
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