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Expression and Perfect Square 2 (Posted on 2015-12-25) Difficulty: 3 of 5
Each of A,B and C is a positive integer such that each of the expressions A*B+A+B, B*C+B+C, C*A+C+A, A*B+C and A*C+B is a perfect square.

(I) Find the smallest value of A+B+C.

(II) Does there exist an infinity of triplets (A,B,C) satisfying the given conditions?
Give reasons for your answer.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re(2): Two small solutions and a conjecture Comment 3 of 3 |
(In reply to re: Two small solutions and a conjecture by Brian Smith)

F(n + 1) F(n- 1) -F(n)^2 = (-1)^n (Cassini)

From which: F(2n + 1) F(2n - 1) = F(2n)^2 +1

We are going to add an irrelevance, because that's what the problem does: let A = F(2k-1)-1, and B = F(2k+1)-1
We have something like: xy=(z^2+1), so (x-1)(y-1) = (z^2+1)-x-y+1, so when we add back A and B in the original equation: (z^2+1)-x-y+1+(x-1)+(y-1) we obtain z^2, after cancelling. If it's true of A and B, it has to be true of B and C, if B = F(2k+1)-1  and C = F(2k+3)-1

Similar manipulation of the other equations produce the squares as required in like manner.

For the others: above we had  (z^2+1)-x-y+1. Now between x and y we have (y-x), so (C+1) = y+(y+(y-x) = 3y-x, giving -2x+2y+z^2+1. But z=(y-x), so 2z+z^2+1 = (z+1)^2

Similar manipulation of the other equations produce the squares as required in like manner.


Edited on December 28, 2015, 9:28 am
  Posted by broll on 2015-12-28 04:10:19

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