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Square Divisiblity (Posted on 2018-06-17) Difficulty: 3 of 5
This problem was inspired by the infamous problem 6 in the 1988 Math Olympiad:

When is a2+b2 divisible by (ab+1)2 where a and b are non-negative integers?

No Solution Yet Submitted by Daniel    
Rating: 3.0000 (1 votes)

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Solution Proof (spoiler) | Comment 2 of 4 |
Consider  (a^2 + b^2) / (ab+1)^2 , which needs to be an integer.

The divisor divides the numerator only if a or > b are 0, in which case (ab + 1)^2 = 0.  Clearly, this always works.

Otherwise, the divisor is bigger than the numerator and the fraction is less than 1

Case 1)  a > 1, b > 1

  (ab+1)^2 = (a^2)(b^2) plus some other positive terms
  Clearly (a^2)(b^2) > (a^2 + b^2), so the divisor is bigger than the numerator

Case 2) a or b = 1

  Without loss of generality, assume a = 1
  Then (ab + 1)^2 = b^2 + 2b +1, which makes the divisor greater
     than the numerator by 2b.

q.e.d.
  



  Posted by Steve Herman on 2018-06-17 20:09:01
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