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Expression and Perfect Square (Posted on 2015-10-10) Difficulty: 3 of 5
Each of X, Y and Z is a nonzero integer such that:

Z = X + Y/X – 1/Y is an integer.

Is Z always a perfect square?
Give reasons for your answer.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution Solution Comment 3 of 3 |
Z = X + Y/X – 1/Y is an integer implies that
Z - X = (Y^2 - X)/(X*Y) is an integer, which then implies that
Z*X - X^2 = (Y^2 - X)/Y is an integer, which then implies that
Z*X - X^2 - Y = -X/Y is an integer.
Then this implies X is a multiple of Y; let X=K*Y.

Then back to the beginning: 
Z = X + Y/X – 1/Y 
Z = K*Y + Y/(K*Y) – 1/Y
Z = K*Y + 1/K - 1/Y
Now to satisfy the requirement for Z to be an integer, 1/K - 1/Y must be zero. (This is where the positive integer requirement speculated by broll comes into play, without that restriction then a nonzero 1/K-1/Y, like if Y=-2 and K=2, is possible.)

Then K=Y, which makes X=Y^2
Back to the beginning again: 
Z = X + Y/X – 1/Y
Z = Y^2 + Y/(Y^2) - 1/Y
Z = Y^2 + 1/Y - 1/Y
Z = Y^2, a perfect square integer.
This verifies Jer's conjecture that X=Z=Y^2 forms the solution set (when the positive integer restriction is in place.)

  Posted by Brian Smith on 2019-03-23 00:55:46
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