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Claim. Theorem 4.3 (p. 320) of O. Ore, Arc coverings of graphs, Ann. Mat. Pura Appl. (4) 55 (1961), no. 1, 315--321, in the corpus's words: a graph on $n$ vertices with at least $\frac12(n-1)(n-2)+2=\binom{n-1}2+2$ edges has a Hamilton circuit; with exactly $\binom{n-1}2+1$ edges the only graphs without one are $K_{n-1}$ with a pendant edge and, for $n=5$, one further graph (the paper's Fig. 3). The count $\binom{n-1}2+2$ is the edge count of Problem 1012 at $k=0$, $\binom{n-1}2+\binom22+1$, and a cycle on $n-0$ vertices is a Hamilton circuit. No graph on $n\le2$ vertices meets the count, so the implication holds for every $n\ge1$ and $f(0)=1$. The page's date is the first day of the issue's month, December 1961 (Crossref).

Covers. The case $k=0$ only, with its sharpness at $\binom{n-1}2+1$ edges. Nothing about $k\ge1$.

Depends on. Nothing in this wiki; the paper's theorem, with its Theorems 3.2 and 4.2, is the whole argument.

Acceptance. Refereed journal publication in the Annali di Matematica Pura ed Applicata (Crossref: issue of December 1961), which is the refereed evidence; Woodall's 1972 paper (p. 749) credits the case $r=0$ to Ore. The site's label SOLVED rests on Woodall, so its credit of $f(0)=1$ to Ore is not listed as reviewed. The proof (pp. 320--321) is followed on the library's result page; nothing here is independently reviewed.