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The unrestricted Lemma 4 on published pp. 7–8 is false as stated. Its conclusion is valid in the following sufficient range. Let q>1q>1 be an integer, k≥1k\ge1, a1,…,ak≥1a_1,\ldots,a_k\ge1 be real, and A=∏iai≤qA=\prod_i a_i\le q. For arbitrary integer did_i there are 1≤T≤q−11\le T\le q-1 and integers di′d_i' such that

Tdi≡di′(modq),∣di′∣≤2qai(Aq)1/k.(1)Td_i\equiv d_i'\pmod q,\qquad |d_i'|\le 2\frac q{a_i}\left(\frac Aq\right)^{1/k}. \tag{1}

No coprimality assumption on the individual steps is required. The application in theorem_2 establishes A≤qA\le q before using this replacement when k≥2k\ge2. Its one-dimensional case uses a unit step directly. The unrestricted printed statement receives no full-proof credit.

Source: published PDF, Lemma 4, pp. 7–8. The restriction and ceiling argument below are compilation-supplied repairs, not an author-issued erratum.

Bears on. Problem 297.

Counterexample to the printed range

Take q=2q=2, k=2k=2, (a1,a2)=(100,1)(a_1,a_2)=(100,1) and (d1,d2)=(1,1)(d_1,d_2)=(1,1). The first bound in (1) would be 2(2/100)100/2=2/5<12(2/100)\sqrt{100/2}=\sqrt2/5<1. But the only allowed multiplier is T=1T=1, and its first residue is odd, so every integer representative has absolute value at least 1. All printed unit assumptions hold. The box count in the printed pigeonhole argument cannot discard integer rounding in this range.

Proof

Put t=(A/q)1/k≤1t=(A/q)^{1/k}\le1, zi=ai/t≥1z_i=a_i/t\ge1 and Ni=⌈zi/2⌉N_i=\lceil z_i/2\rceil. For z≥1z\ge1,

⌈z/2⌉≤z,with strict inequality if z>1.\lceil z/2\rceil\le z, \quad\text{with strict inequality if }z>1.

At z=1z=1 there is equality; for 1<z≤21<z\le2 the ceiling is 1; for z>2z>2, ⌈z/2⌉<z/2+1<z\lceil z/2\rceil<z/2+1<z. Since ∏izi=q>1\prod_i z_i=q>1, at least one factor has strict inequality. Consequently ∏iNi<q\prod_i N_i<q.

Partition [0,q)[0,q) in coordinate ii into NiN_i half-open intervals of equal length. Their length is at most 2qt/ai2qt/a_i. For u=0,…,q−1u=0,\ldots,q-1, form the vector of least nonnegative residues of udiud_i. Two of these qq vectors lie in the same product box. If their indices are u≠vu\ne v, let TT be the nonzero residue of u−vu-v in {1,…,q−1}\{1,\ldots,q-1\}, and let di′d_i' be the difference of the two representatives in coordinate ii. These differences have the required congruences and absolute values strictly less than the indicated interval lengths. This proves (1).