Common Contracts
1 similar null contracts
Licence agreement concerning inclusion of doctoralNovember 16th, 2009
To count Q2(x), we observe that since (2d + 1)2 4c = 0, we need (2d + 1)2 n2 + 4c x6/4 4x, which can only happen if d 1 x3/4 x1/4 1. So when d is large enough to get a solution, the difference between (2(d + 1) + 1)2 and (2d + 1)2 is at least 4x3/4 8x1/4, which is greater than x3/4, for x sufficiently large. Around every multiple of n2, we have an interval of length 8x in which (2d + 1)2 must lie for solutions to occur. The number of d that can lie in such an interval is at most 8x/x3/4 + 1 = 8x1/4 + 1 and the number of intervals is at most x2/n2 + 1 x2/4 + 1.
