Chowla’s conjecture.

*(English)*Zbl 1154.11339From the author’s introduction: Let \(K=\mathbb Q(\sqrt(d))\), where \(\mathbb Q\) is the rational field and \(d\) is a square-free positive integer, and let \(h(d)\) be the class number of this field. In [S. Chowla and J. Friedlander, Glasg. Math. J. 17, 47–52 (1976; Zbl 0323.12006)], S. Chowla conjectured that \(h(4p^2+1)>1\) if \(p>13\) is an integer, which is proved to be true in this paper. The work here has its origins in the author’s paper [Acta Arith. 106, No. 1, 85–104 (2003; Zbl 1154.11338)], in which he established a conjecture of H. Yokoi that \(h(p^2+4)>1\) for \(p>17\). In fact, essentially the same proof works with appropriate modifications. Note that Siegel’s theorem tells us that the class number is greater than 1 once \(p\) is sufficiently large, in both cases; however, Siegel’s theorem does not indicate what “sufficiently large” means. Here the author determines that using a quite different method. His main result is as follows.

Theorem. If \(d\) is square-free, \(h(d)=1\) and \(d=4p^2+1\) with some positive integer \(p\), then \(p\) is a square for at least one of the following moduli: \(q=5,7,41,61,1861\) (that is, \((d/q)=0\) or 1 for at least one of the listed values of \(q\)).

Combining this with Fact B (which implies that if \(h(d)=1\), then \(d\) is a quadratic nonresidue modulo any prime \(r\) with \(2<r<p\)) he obtains:

Corollary. If \(d\) is square-free, and \(d=4p^2+1\) with some integer \(p>1861\), then \(h(d)>1\).

What concerns the small solutions, in the same way as in the author’s cited paper, he can easily prove (see Section 2) that \(h(4p^2+1)>1\) if \(13<p\leq 1861\). Hence Chowla’s conjecture follows. He searches these final few cases to show that \(h(4p^2+1)=1\) only for \(p=1,2,3,5,7,13\).

The main lines of the proof are the same as in the author’s cited paper, but some modifications are needed; the most significant modifications can be found in the statement and proof of Lemma 1. The present proof also requires computer work.

Theorem. If \(d\) is square-free, \(h(d)=1\) and \(d=4p^2+1\) with some positive integer \(p\), then \(p\) is a square for at least one of the following moduli: \(q=5,7,41,61,1861\) (that is, \((d/q)=0\) or 1 for at least one of the listed values of \(q\)).

Combining this with Fact B (which implies that if \(h(d)=1\), then \(d\) is a quadratic nonresidue modulo any prime \(r\) with \(2<r<p\)) he obtains:

Corollary. If \(d\) is square-free, and \(d=4p^2+1\) with some integer \(p>1861\), then \(h(d)>1\).

What concerns the small solutions, in the same way as in the author’s cited paper, he can easily prove (see Section 2) that \(h(4p^2+1)>1\) if \(13<p\leq 1861\). Hence Chowla’s conjecture follows. He searches these final few cases to show that \(h(4p^2+1)=1\) only for \(p=1,2,3,5,7,13\).

The main lines of the proof are the same as in the author’s cited paper, but some modifications are needed; the most significant modifications can be found in the statement and proof of Lemma 1. The present proof also requires computer work.

Reviewer: Olaf Ninnemann (Berlin)