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Divisibility to Square II (Posted on 2016-08-07) Difficulty: 4 of 5
This is a followup to Divisibility to square.

Each of x and y is a positive integer such that x^2 + y^2 + x is divisible by xy.

1: Prove that there is an infinite number of (x,y) which make the quotient equal to 3.

2: Prove or disprove 3 is the only integer quotient possible.

No Solution Yet Submitted by Brian Smith    
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Part 1, offered without proof | Comment 1 of 2
Given F(n) = nth term in the Fibonacci sequence 1,1,2,3,5,8,13,21,34,55,89,144,...

then x=(F(2n+1))^2 is a solution when either y=(F(2n))^2 + 1 or y=(F(2n+2))^2 + 1.

For example, F(7)=13, so (169,65) and (169,442) are solutions. 



  Posted by xdog on 2016-08-07 13:19:05
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