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On a conjecture of Rudin on squares in arithmetic progressions. (English) Zbl 1314.11059

Let \(q\neq 0\) and \(a\) be integers, and \[ Q(N; q,a):=\#\{ qi + a : i =0,1,\dots,N-1 \} \cap \{ x^2 : x \in \mathbb Z \} \] and let \(Q(N):=\max\{ Q(N;q,a) : q,a\in\mathbb Z,\;q\neq 0\}\). For example, recall Fermat’s result: there are no four consecutive terms of a non-constant arithmetic progression that are squares, and it can be formulated as \(Q(4) < 4\); in fact, \(Q(4)=3\). P. Erdős conjectured in [Monographies Enseign. Math. 6, 81–135 (1963; Zbl 0117.02901)] that \(Q(N) = o(N)\), and it was proved by E. Szemerédi in [Stud. Sci. Math. Hung. 9, 417 (1976; Zbl 0318.10029)]. In fact W. Rudin conjectured in [J. Math. Mech. 9, 203–227 (1960; Zbl 0091.05802)] that \(Q(N)=O(\sqrt N)\). E. Bombieri et al. proved in [Duke Math. J. 66, No. 3, 369–385 (1992; Zbl 0771.11034)] that \(Q(N)=O(N^{2/3 + o(1)} )\), and U. Zannier and E. Bombieri improved it in [Atti Accad. Naz. Lincei, Cl. Sci. Fis. Mat. Nat., IX. Ser., Rend. Lincei, Mat. Appl. 13, No. 2, 69–75 (2002; Zbl 1072.11010)] to \(Q(N)=O(N^{3/5 + o(1)} )\). Strong Rudin’s conjecture is known as \[ Q(N)=Q(N; 24,1)=\sqrt{ \frac 83\, N} + O(1) \] for \(N \geq 6\), and the authors of the paper under review computationally verify that \(Q(N)=Q(N; 24,1)\) for \(6 \leq N \leq 52\).
It is clear that \(Q(N) - Q(N-1) \leq 1\), and the authors notice that among \(\{ 6,\dots, 52\}\), we have \(Q(N-1)+1=Q(N)\) only for \(N \in \{8,13,16,23,27,36,41,52\}\), and that \(N-1\) are consecutive terms of the generalized pentagonal numbers. They prove that if \(N \in \{8,13,16,23,27,36,41,52\}\) and \(Q(N)=Q(N;q,a)\) such that \(\gcd(q,a)\) is square free and \(q>0\), then \(q=24\) and \(a=1\). Since \(qi+a\) being a square implies \((q\ell^2)i + a\ell^2\) being a square, their result proves that the arithmetic progression \(24i + 1\) is essentially a unique one with the maximal property when \(Q\)-values jump. Motivated by this uniqueness result they introduce a super-strong Rudin’s Conjecture: In addition to Strong Rudin’s Conjecture, if \(Q(N)=Q(N;q,a)\) such that \(\gcd(q,a)\) is square free, \(q>0\), and \(N-1\geq 7\) is a generalized pentagonal number, then \(q=24\) and \(a=1\).
Let us briefly introduce their approach. Let \(I\) be an index set (for multiple arithmetic progressions), and let \(Z_I\) be the set of pairs \((q,a)\) such that \(\gcd(q,a)\) is square free, \(q\neq 0\), and \(qi+a\) is a square for all \(i\in I\). Given \(N\), if \(Z_I\) is empty for all subsets \(I\) of \(\{ 0,1,\dots,N\}\) with \(\#I =Q(N)+1\), then \(Q(N+1)=Q(N)\), and otherwise, \(Q(N+1)=Q(N)+1\). To determine whether \(Z_I\) is empty, they construct a projective curve \(C_I\) for each \(I\) with \(k+1\) elements. The genus of \(C_I\) is \((k-3)2^{k-2}+1\), and \(C_I(\mathbb{Q})\) minus \`\` trivial points\'\'is naturally bijective to \(Z_I\). If \(\#I = 4\), \(C_I\) is an elliptic curve, and if \(\#I = 5\), \(C_I\) is a curve of genus \(5\). The authors employ various methods from the theory of elliptic curves and curves of higher genus to determine whether \(Z_I\) is empty.

MSC:

11N13 Primes in congruence classes
11G30 Curves of arbitrary genus or genus \(\ne 1\) over global fields
11B25 Arithmetic progressions
11D45 Counting solutions of Diophantine equations
14H25 Arithmetic ground fields for curves

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References:

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