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III
                        The Erd˝ os–Fuchs Theorem












                        There has always been some fascination with the possibility of near
                        constancy of the representation functions r i (n) (of I (7), (8) and (9)).
                        In Chapter I we treated the case of r + (n) and showed that this could
                        noteventuallybeconstant.Thefactthatr(n)cannotbeconstantforan
                        infinite set is really trivial since r(n) is odd for n   2a, a ∈ A, and
                        even otherwise. The case of r − (n) is more difficult, and we will treat
                        it in this chapter as an introduction to the analysis in the Erd˝ os–Fuchs
                        theorem.
                           The Erd˝ os–Fuchs theorem involves the question of just how nearly
                        constant r(n) can be on average. Historically this all began with the
                                   2
                        setA  {n : n ∈ N 0 },thesetofperfectsquares,andtheobservation
                        that then  r(0)+r(1)+r(2)+···+r(n)  , the average value, is exactly equal to
                                         n+1
                          1  times the number of lattice points in the quarter disc x, y ≥ 0,
                         n+1
                          2
                               2
                        x + y ≤ n. Consideration of the double Riemann integral shows
                        that this average approaches the area of the unit quarter circle, namely
                                                    r(0)+r(1)+r(2)+···+r(n)  π
                        π/4, and so for this set A,                    →      (r(n) is on
                                                           n+1              4
                        average equal to the constant π/4.)
                           The difficult question is how quickly this limit is approached. Thus
                        fairly simple reasoning shows that


                              r(0) + r(1) + r(2) + ··· + r(n)      π          1
                                                                      + O    √     ,
                                           n + 1                   4           n
                        whereas more involved analysis shows that


                              r(0) + r(1) + r(2) + ··· + r(n)      π          1
                                                                      + O          .
                                           n + 1                   4         n 2/3

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