Another case of divisibility (Posted on 2009-02-15)

	Submitted by pcbouhid
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See problem 4_eureka3. If I = (a2b + a + b) / (ab2 + b + 7) is integer, then [b(a2b + a + b) - a(ab2 + a + b)] / (ab2 + b + 7) = (b2 - 7a) / (ab2 + b + 7) is integer. Since (b2 - 7a) < b2 < ab2 + b + 7, then we have N = (b2 - 7a) / (ab2 + b + 7) < 1. If N = 0, then b2 = 7a, and so b is a multiple of 7 (say b = 7k), and (7k)2 = 7a give us a = 7k2. So all the pairs (a, b) = (7k2, 7k) for k positive integer, satisfy the conditions of the problem. If N < 0, we must have b2 < 7a and N =< -1 (since it must be integer), and so 7a > (7a - b2) >= ab2 + b + 7. Then 7a > ab2 ===> b2 < 7 ===> b = 1 or 2. If b = 1, N = (1 - 7a) / (a + 8) = - 7 + 57/(a + 8) and so, (a + 8) must divide 57. Thus (a + 8) = 19 .... a = 11, or (a + 8) = 57 ..... a = 49. For b = 1: If a = 11 we have I = 133 / 19 = 7, and if a = 49, we have I = 2451 / 57 = 43. For b = 2, N = (4 - 7a) / (4a + 9). Since (4 - 7a) > - 18 - 8a = -2(4a + 9), if N is a negative integer, we must have N = - 1, which leads to a = 13/3, that doesn´t satisfy the conditions of the problem. So the pairs (a, b) are (7k2, 7k) - k positive integer -, (11, 1) and (49, 1).

	Subject	Author	Date
	Alternative Methodology	K Sengupta	2009-02-15 14:33:50
	Analytic Solution	K Sengupta	2009-02-15 14:31:13
	partial list -- possible spoiler.	Charlie	2009-02-15 14:19:34