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Claim. Problem 545 asks, for a graph $G$ with $m=\binom n2+t$ edges ($0\le t<n$) and no isolated vertices, whether $R(G)\le R(H)$, where $H$ is $K_n$ with a new vertex joined to $t$ of its vertices and $R$ is the diagonal two-color Ramsey number. The comments posted to the site's thread on 28 October 2025 under the account LouisD give the matching $mK_2$ as a counterexample for $m\in\{3,4,5,7,8,9\}$. The matching has $R(mK_2)=3m-1$ (Cockayne and Lorimer, 1975, as the comments cite it; the upper bound by removing a red edge meeting a blue one and inducting, the lower bound from a set of $m-1$ vertices whose edges are all red against $2m-1$ others). Against it: for $m=3$, $H=K_3$ and $R(3K_2)=8>6=R(K_3)$; for $m=4$ and $m=5$, $H$ is a subgraph of $K_4-e$ and equal to it, with $R(K_4-e)=10$ from Radziszowski's survey, below $R(4K_2)=11$ and $R(5K_2)=14$; for $m=7$, $H=K_4^{+1}$ is $K_4$ with one pendant edge, and a written argument in the thread gives $R(K_4^{+1})=18=R(K_4)$, below $R(7K_2)=20$; for $m=8$ and $m=9$, $H$ is a subgraph of $K_5-e$ and equal to it, with $R(K_5-e)=22$ from the same survey, below $R(8K_2)=23$ and $R(9K_2)=26$. The same matching fails the statement at $m=2$, the smallest case the site's commentary lists: $m=2=\binom22+1$ gives $n=2$, $t=1$ and $H=P_3$, the path with two edges, with $R(P_3)=3$ (two of the three edges of $K_3$ share a color and share a vertex), while $R(2K_2)=5$, the case $m=2$ of $R(mK_2)=3m-1$ (a red triangle in $K_4$ with the three edges at the fourth vertex blue has no two disjoint edges of one color; in any $2$-coloring of $K_5$ one color has at least five edges, more than a star or a triangle on five vertices can hold, so that color contains two disjoint edges). A universal statement with a false instance is false, so the question as the site states it, which quantifies over every $m$, has the answer no. The comments note that the comparison holds at $m=6$ ($R(6K_2)=17<18=R(K_4)$), which the site's curator confirms from Burr's 1989 table of the Ramsey numbers of graphs with at most six edges, and that the matching gives no counterexample at $m=10$ ($R(10K_2)=29<43\le R(K_5)$); the commenter asks whether the statement holds from ten edges on.

The instance $m=2$ was posted to the same thread later the same day under the account Adenwalla, in a comment the site marks as addressed, and a comment of the next day under a third account adds that $m=1$ allows no comparison; the site's acknowledgment line thanks all three accounts. The curator's commentary credits the small-$m$ failures, $m=2$ among them, to the account named by this page, which posted first, so the later comment is disclosed here rather than given a page of its own. The problem page verifies the two Ramsey numbers of the $m=2$ case without a source, an authored check that warrants nothing here.

Standing. Rejected: answers the site's wording, not the corrected statement. The counterexamples lie at $m\le9$ and settle no instance of the question for all sufficiently large $m$ that the problem page shows; the comments themselves ask how large is large enough. The site's curator, T. F. Bloom, wrote the failures into the problem's commentary, crediting the account by name and stating that the statement fails for $2\le m\le5$ and $7\le m\le9$ (page last edited 2 December 2025), added the contributors to the page's acknowledgment line, marked the comments as addressed, and kept the label OPEN for the large-$m$ question. The tabulated values the argument uses, $R(mK_2)=3m-1$, $R(K_4-e)=10$ and $R(K_5-e)=22$, are classical and cited in the thread from refereed sources; the bound $R(K_4^{+1})\le18$ rests on the thread's written argument. There is no entry on the site's proof-claim tab.

Depends on. Nothing in this wiki; the argument is the comments' own comparison of tabulated Ramsey numbers.