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III. The Erd˝ os–Fuchs Theorem
38
At last, combining (19), (20), and (21) allows the conclusion
C e 1
2
≤ N 1 − r log + √ . à 22)
M 1 − r 2 N(1 − r )
2 α
Once again we are masters of the parameters (subject to (18)),
√
2 1
and so we elect to choose r, so that N 1 − r √ . Thus
2 α
N(1−r )
3
our choice is to make 1 2 N 2α+1 and note happily that our side
1−r
condition (18) is satisfied. Also “plugging” this choice into (22) gives
C 4α−1
≤ N 4α+2 (2 + 3 log N). (23)
M
Well, success is delicious. We certainly see in (23) the fact that
1 4α−1
α ≥ . (If the exponent of N, , were negative then this right-
4 4α+2
hand side would go to 0, 2+3 log N notwithstanding, and (23) would
become false for large N.)