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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)

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Some Thoughts Two small solutions and a conjecture | Comment 1 of 3
To start I rearranged the first three expressions as:
(A+1)*(B+1) = X^2+1
(A+1)*(C+1) = Y^2+1
(B+1)*(C+1) = Z^2+1

Let P = (A+1)*(B+1)*(C+1).  Then solving for A, B and C:
A = sqrt[P]/(Z^2+1) - 1
B = sqrt[P]/(Y^2+1) - 1
C = sqrt[P]/(X^2+1) - 1

Then I sifted through numbers of the form x^2+1 until I got P to be a square.  The smallest nontrivial case came from X=3, Y=5, Z=8.  This yielded A=1, B=4, C=12.  A*B+C = 4^2, A*C+B = 4^2 and B*C+A = 7^2.  Then A+B+C = 17.

The next set I found came from X=8, Y=13, Z=21 yielding A=4, B=12, C=33.  Then A*B+C = 9^2, A*C+B = 12^2, and B*C+A = 20^2.  Then A+B+C = 49.

My conjecture is that alternating triplets from the Fibonacci sequence always produce {X, Y, Z} such that the triplet {A, B, C} are solutions to the problem.  

Trying the next such triplet {X, Y, Z} = {21, 34, 55}.
Then {A, B, C} = {12, 33, 88}.  Then A*B+C = 22^2, A*C+B = 33^2, and B*C+A = 54^2.

  Posted by Brian Smith on 2015-12-25 11:42:52
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