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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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Supplementary questions | Comment 4 of 7 |

There are two traps here for the unwary analyst:
(a) q is a quotient and r a remainder, thus 0<=r<q
(b) the main one is that there is no convenient way of producing the quotient and remainder of (x^2+y^2)/(x+y) algebraically.

 So it helps to start at the other end of the problem:
1. By inspection, 0<q<45, since 45^2=2025
2. r must be less than q as it is a remainder (see (a));
so q^2+r=1977, 0<r<q<(1977)^(1/2): q=44,r=41
3. (x^2+y^2)/(x+y)=44+41/(x+y)
4. (x-44)x+(y-44)y = 41
5. {x,y}={-6,7}{-6,37}{7,-6}{7,50}(37,-6}{37,50}{50,7}{50,37}
of which {{7,50} and {37,50} meet the stipulation of the problem.

Thus, the problem only 'works' in years where the remainder is less than the quotient.  Given that 45^2-44^2=89, no parallel problem has been available since 1980(44^2+44=1980). The problem will next 'work' again in 2025 when {q,r} = {45,0}

Edited on February 9, 2011, 5:52 am
  Posted by broll on 2011-02-09 05:50:57

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