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Claim. For every fixed $c>3/4$ and every sufficiently large prime $p$, every interval $(n,n+p^c)$ with $n\geq0$ contains integers $a,b$ with $ab\equiv1\pmod p$. This answers yes the question of Problem 445 for the exponents $c>3/4$. The source is T. D. Browning and A. Haynes, Incomplete Kloosterman sums and multiplicative inverses in short intervals, Int. J. Number Theory 9 (2013), no. 2, 481–486, first posted as arXiv:1204.6374 on 28 April 2012. Its Theorem 1 states: for subintervals $I_1^{(j)},I_2^{(j)}\subseteq(0,p)$, $1\le j\le J$, of lengths $H$ and $K$ with the $I_1^{(j)}$ pairwise disjoint, some $j$ has $x\in I_1^{(j)}$, $y\in I_2^{(j)}$ with $xy\equiv1\pmod p$ once $J\gg p^3(\log p)^4/(H^2K^2)$. The paper notes that $J=1$ recovers the two-interval condition $HK\gg p^{3/2}(\log p)^2$, which it attributes to Heath-Brown's 2000 article; its proof runs through a mean value theorem for incomplete Kloosterman sums (its Theorem 2) and Weil's bound. The problem page records the short deduction: the integers of $(n,n+p^c)$ reduce modulo $p$ to at most two blocks of consecutive nonzero residues, the longer of which has at least $(p^c-O(1))/2$ elements, and for $3/4<c<1$ the product of two such lengths exceeds $Cp^{3/2}(\log p)^2$ once $p$ is large, uniformly in $n$; the case $c\geq1$ follows from $c=7/8$ by inclusion.

Covers. Every fixed exponent $c>3/4$, for every translate $n\geq0$. Not covered: the range $1/2<c\leq3/4$, which the problem asks about and which remains open; the logarithmic factor in the criterion excludes $c=3/4$ itself.

Credit. The site's remark credits the range $c>3/4$ to Heath-Brown, and Browning and Haynes credit the two-interval bound to Heath-Brown's article Arithmetic applications of Kloosterman sums, Nieuw Arch. Wiskd. (5) 1 (2000), 380–384. That article displays the count for the origin box $1\le m,n\le M$ only, for a general residue, which for the problem's residue $1$ settles no instance; the problem page records it in its Current assessment. The site's remark also credits Heilbronn, without a publication, with the case of $c$ sufficiently close to $1$.

Acceptance. Refereed: International Journal of Number Theory 9, no. 2 (2013), 481–486. The site labels the problem OPEN, so its remark crediting the range $c>3/4$ is not an acceptance, and no reviewed evidence is listed.

Depends on. Theorem 1 of Browning and Haynes with the short deduction stated under Claim.