Problem 445

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claims/: The 1 claim page of Problem 445, one per claimant's result; the problem's standing derives from them.

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Statement. Is it true that, for any $c>1/2$, if $p$ is a sufficiently large prime then, for any $n\geq 0$, there exist $a,b\in(n,n+p^c)$ such that $ab\equiv 1\pmod{p}$?

Status. Open. The site's remark (page last edited 27 December 2025) credits Heilbronn, unpublished, with the case of $c$ sufficiently close to $1$ and Heath-Brown with every $c>3/4$. The range $c>3/4$ is recorded as an accepted partial claim on the Browning and Haynes claim page, whose refereed two-interval criterion states the bound that the site and Browning and Haynes credit to Heath-Brown. The standing in the frontmatter derives from the claim pages.

Source. erdosproblems.com/445, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #445, https://www.erdosproblems.com/445.

References.

• [He00] Heath-Brown, D. R., Arithmetic applications of Kloosterman sums. Nieuw Arch. Wiskd. (5) 1 (2000), no. 4 (December 2000), 380-384, a write-up of a Kloosterman centennial lecture, online.

Formalization. Statement in formal-conjectures.

Current assessment

The exact question is open for $1/2<c\leq3/4$. The range $c>3/4$ is settled for every translate $n$ by the Kloosterman-sum method, through Browning and Haynes's 2013 criterion for arbitrary intervals, recorded on the Browning and Haynes claim page as an accepted partial claim with refereed evidence. Heath-Brown's 2000 article has no claim page of its own: it displays the count of solutions of $mn\equiv a\pmod p$ in the origin box $1\le m,n\le M$ for a general residue $a$, and for the problem's residue $1$ the origin case is trivial, since $a=b=1$ lies in $(0,p^c)$, so the display settles no instance of the problem; the site's remark and Browning and Haynes credit him with the two-interval bound, which the claim page states in their form. Heath-Brown's article calls improving the exponent $3/4$ an open problem. The search scope is the site page, the two primary papers, Browning's publication list and a web literature search for a theorem below the exponent $3/4$, carried out on 2026-09-05 and 2026-09-06; it found no such theorem and no proof claim. Noncoverage by a partial theorem alone does not establish that the remaining range is open, and no later search is recorded.

The proofs of the Heath-Brown estimate and of the Browning–Haynes criterion are not transcribed in the library; the claim page rests on the refereed publication of the criterion, not on a proof review by this corpus.

Known Results

Erdős and Graham, Old and new problems and results in combinatorial number theory (1980), printed p. 89, state the translated-interval question and attribute the case with $c$ sufficiently close to $1$ to Heilbronn, whose proof is unpublished.

Heath-Brown, Arithmetic applications of Kloosterman sums (2000), printed p. 382, gives an origin-rectangle estimate. For a prime $p\geq3$, $p\nmid a$ and $1\leq M\leq p$ (a range the print's derivation needs but leaves implicit), writing

$$N_a(M)=\#\{(m,n):1\leq m,n\leq M,\ mn\equiv a\pmod p\},$$

the completion lemma for incomplete exponential sums and Weil's bound $|S(m,k;p)|\leq2p^{1/2}$ for $p\nmid k$ give the displayed error bound

$$\left|N_a(M)-\frac{M^2}{p}\right| \leq2(\log p)(1+\log p)p^{1/2} <4(\log p)^2p^{1/2}.$$

Consequently the least such scale satisfies $M(a)\leq2(\log p)p^{3/4}$. The asymptotic count follows, for example, when $M^2/[p^{3/2}(\log p)^2]\to\infty$. This displayed source passage concerns the positive origin rectangle.

Browning and Haynes, Incomplete Kloosterman sums and multiplicative inverses in short intervals, arXiv:1204.6374v1 (28 April 2012), pp. 1–2, state that arbitrary subintervals $I_1,I_2$ of $(0,p)$ contain integers $x,y$ with $xy\equiv1\pmod p$ whenever

$$|I_1||I_2|\geq C p^{3/2}(\log p)^2$$

for a sufficiently large absolute constant $C$. Theorem 1 on p. 2 recovers this criterion by setting $J=1$. The article appeared in International Journal of Number Theory 9 (2013), 481–486, DOI 10.1142/S1793042112501448.

Here is the short application to the exact translated question. Fix $3/4<c<1$. The integers in $(n,n+p^c)$ form a block of $p^c+O(1)$ consecutive integers, uniformly in $n$. Reduce modulo $p$, delete residue zero, and take the longer of the at most two nonwrapping components. It contains at least $(p^c-O(1))/2$ nonzero consecutive residues. Using that component for both intervals, the ratio of their size product to $p^{3/2}(\log p)^2$ tends to infinity because $2c-3/2>0$. The criterion produces an inverse pair whose representatives lie in the original open interval. All bounds are uniform in $n$. For $c\geq1$, apply the established case $c=7/8$ and interval inclusion. Thus every fixed $c>3/4$ is covered. The logarithmic factor prevents this argument from including $c=3/4$.

The short application above is an existing-source deduction, not a solution of the remaining range.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.

• browning_2013_incomplete_kloosterman_sums
• browning_2013_incomplete_kloosterman_sums / corollary
• browning_2013_incomplete_kloosterman_sums / theorem_1
• browning_2013_incomplete_kloosterman_sums / theorem_2
• heathbrown_2000_arithmetic_applications_kloosterman_sums
• heathbrown_2000_arithmetic_applications_kloosterman_sums / estimate_p382
• heathbrown_2000_arithmetic_applications_kloosterman_sums / lemma_p380
• erdos_1980_old_new_problems_results_combinatorial_number_theory