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A 2547 Puzzle (Posted on 20070329) 

a and b are positive integers. Dividing a^{2} + b^{2} by a + b we obtain the quotient as q and the remainder as r.
Determine analytically all possible pairs (a, b) such that q^{2} + r = 2547

Submitted by K Sengupta

Rating: 2.0000 (2 votes)


Solution:

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a^2 + b^2 >= (a + b)^2/2, so q >= (a + b)/2. Hence r < 2q.
The largest square less than 2547 is 2500 = 50^2 and:
2547 = 50^2 + 47.
The next largest gives 2547 = 49^2 + 146 . But 146 > 2*43.
So we must have q = 50, r = 47.
Hence a^2 + b^2 = 50(a + b) + 47
So, (a  25)^2 + (b  25)^2 = 1297 = 36^2 + 1^2
Accordingly, (a25, b25) = (36, 1); (36, 1); (1, 36); (1, 36), since both a and b are positive integers.
Consequently, (a,b) = (61, 26); (61, 24); (26,61); (24, 61)

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