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Q^2 + R = 1977 (Posted on 2011-02-08) Difficulty: 3 of 5
Each of X and Y is a positive integer with X ≤ Y. The quotient and the remainder obtained upon dividing X2 + Y2 by X+Y are respectively denoted by Q and R.

Determine all possible pairs (X, Y) such that Q2 + R = 1977

Supplementary questions:

This problem has been out of circulation for quite some time. Why? When is it likely to come back into favour?

No Solution Yet Submitted by K Sengupta    
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re: Supplementary questions | Comment 5 of 7 |
(In reply to Supplementary questions by broll)

"(a) q is a quotient and r a remainder, thus 0<=r<q"

I don't understand this.  If 13 is divided by 5, the quotient is 2 and the remainder is 3, a number larger than the quotient. The remainder must be smaller than the divisor, but that's not what you're saying.

Or, in particular, when

x=31  y=53          dividend=3770  divisor=84        quotient =44  remainder=74

and the value arrived at is 2010.  The remainder,74, is, as appropriate, smaller than the divisor, 84, but need not be smaller than the quotient, 44.

Edited on February 9, 2011, 12:10 pm
  Posted by Charlie on 2011-02-09 11:52:58

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