Each of X and Y is a positive integer with X ≤ Y. The quotient and the remainder obtained upon dividing X^{2} + Y^{2} by X+Y are respectively denoted by Q and R.
Determine all possible pairs (X, Y) such that Q^{2} + 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?
(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 20110209 11:52:58 