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Source. U. V. Linnik, “On Erdös's theorem on the addition of numerical sequences,” first through fourth lemmas, printed pp. 68–70 (PDF pp. 2–4). The English proof in the twelve-page MathNet scan is controlling; see [[integer_sequences/linnik_1942_erdos_theorem_addition_numerical_sequences/_index|the source index]] for the version and its discrepancies. The construction uses the sufficient versions proved below, with its changes recorded in [[integer_sequences/linnik_1942_erdos_theorem_addition_numerical_sequences/theorem|the main result]].
Write . For a finite set of integers, its Weyl sum is , and its value at zero is . Pointwise absolute values of sums and cardinalities of sets have their usual meanings; they are not interchanged. All logarithms are natural.
First lemma: the printed statement and a sufficient specialization
Linnik's first lemma, p. 68, asserts the following. Put . If , , is an integer,
where are coprime integers, , and , then
The paper calls this an immediate consequence of Vinogradov's Theorem 1. The full explicit numerical range in this printed assertion is recorded here as a source claim; it is not independently established by the calculation below. In particular, replacing in its right-hand side by the number of terms is not justified. The construction needs only the following narrower, normalized statement.
Sufficient Weyl estimate. There is an absolute such that the following holds for integers :
If , , , and , then, with ,
External input. The two relevant ranges of Vinogradov's Theorem 1, as quoted on Linnik's p. 68, specialize to the following for a polynomial of degree , leading coefficient , and a sum over consecutive integers starting after a positive integer. Put . For the multiplier they give
and
In case 3) of Linnik's quotation the upper exponent is printed as . It is read here as , with ; renaming the degree as then gives the cutoff in (V3). The proof below uses (V3) only for .
The multiplier restrictions in the quoted theorem are automatic for . The external source is I. M. Vinogradov, “Estimations of trigonometrical sums,” Bulletin de l'Académie des Sciences de l'URSS, nos. 5–6 (1938), 505–524, Theorem 1, Linnik's reference 4. Its proof is an external dependency, not reproduced here.
Proof of (W). Write . Since ,
Uniformly for , sufficiently large satisfies
Consequently . If , apply (V2). Otherwise set . Then , so (V3) applies: for . In both cases and
For all sufficiently large , the hypotheses give , , and . Therefore
The last inequality holds eventually because . Choose one absolute beyond all the thresholds used above. This proves (W), uniformly in , , and the admissible rational approximation. Conjugating the sum gives the identical bound for the negative phase.
Second lemma: exceptional primes for residue counts
Let be a set of distinct integers in , where , and let
Among the primes , let count those for which more than residue classes contain at most elements of . There is an absolute constant such that
External input. Use the large-sieve inequality in its absolute-constant form. If points on are separated by at least , then
Linnik invokes the method of his “The large sieve,” C. R. U. R. S. S. (listed as “in print” in reference 5), with . The large-sieve theorem is the external dependency; the deduction of (1.2) is given here.
Proof. For a bad prime , let be the residue counts and let of them be at most . The total in the other classes is at least . There cannot be low classes, since their total would be at most . Cauchy–Schwarz gives
For the last inequality, the function is increasing for , and its value at exceeds . Finite Fourier orthogonality now gives
The nonzero fractions over all these primes are distinct and -separated on the circle. Summing and applying (LS) yields
because and . Absorb the absolute constants into .
Third lemma: a prime with uniform representation counts
Let be finite sets of distinct integers, with cardinalities for a fixed . Let be a positive integer with
For all , where the threshold depends only on , some prime satisfies, for every ,
The printed lemma (p. 69) states this for sums only. Section 3 (p. 72) applies it to the differences , and the proof below also gives the same conclusion for , with a possibly relabeled target residue.
Proof. By the second lemma, fewer than
primes are bad for at least one of the sets. The prime number theorem, the additional external input used on p. 69, gives at least primes in for large . The ratio of the first bound to this lower bound tends to zero, uniformly in the allowed , since . Thus there is a prime good for both sets.
For each set, at most residue classes have count at most . Fix . Of the pairs of residues , at least avoid both exceptional sets. Each such pair contributes at least ordered pairs of elements. This proves the weaker constant printed in (1.3).
For differences, use the residue pairs in the same count. The same prime works, with the same exceptional-class bound and the same constant, for every difference residue.
Fourth lemma: a corrected finite-cutoff alternative
As printed (pp. 69–70), the lemma takes a sequence of positive density at least , with counting function , fixed numbers , and , and the sequence of the terms of and their successors, with counting function . Its alternative is that either (1, 4), or for every term of up to the numbers belong to up to , with at most exceptions. The conclusion writes for the of the hypothesis, and Section 3 applies the lemma with (p. 71).
The assertion so printed assumes only . That is insufficient when the conclusion requires all successors to remain below the cutoff. For example, take , , and . Then gains no points below , while ten starting points fail the successor condition, more than .
The following additional threshold is sufficient and is all that the construction needs. Let , , , and let be an integer with
Put , , and
Then either
or fewer than elements fail the condition
No density hypothesis is needed for this finite combinatorial assertion.
Proof. At most starting points lie in the terminal strip . For every other failing start , take the least with . By minimality, , so is one of the new elements of below . A new element can be charged by at most starts, all lying among its predecessors. Thus the number of failing starts is at most . If the first alternative fails, this is less than
This proves the corrected alternative, including the terminal boundary and the multiplicity of the charging map.
Bears on. Problem 38, only as inputs to Linnik's essential-component construction on [[integer_sequences/linnik_1942_erdos_theorem_addition_numerical_sequences/theorem|the main result page]]; none of these lemmas concerns the problem on its own.