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Statement

Following the remark on p. 116, let $N_m$ be the largest number $N$ for which $1,2,\ldots,N$ can be distributed into $m$ rows so that no row contains the difference of two of its numbers (finite by the Hilfssatz). No row may hold both $x$ and $2x$, since $2x-x=x$; so $N_m$ is the Schur number $S(m)$ of the literature, the largest $N$ with a partition of $\{1,\ldots,N\}$ into $m$ classes free of solutions of $a+b=c$, $a=b$ allowed.

Construction (pp. 116--117). If rows $x_1,x_2,\ldots;\ \ldots;
u_1,u_2,\ldots$ distribute $1,\ldots,N_m$ with the property, then the $m+1$ rows

$$3x_1,\ 3x_1-1,\ 3x_2,\ 3x_2-1,\ \ldots;\quad\ldots;\quad 3u_1,\ 3u_1-1,\ 3u_2,\ 3u_2-1,\ \ldots;\quad 1,\ 4,\ 7,\ \ldots,\ 3N_m+1$$

distribute $1,\ldots,3N_m+1$ with the property; the paper asserts this "wie man leicht erkennt" and illustrates it for $m=2$ (p. 117), passing from the rows $1,4$ and $2,3$ to the rows $3,2,12,11$; $6,5,9,8$; $1,4,7,10,13$.

Conclusion (p. 117). Hence $N_{m+1}\ge3N_m+1$, and since $N_1=1$,

$$N_m\ \ge\ 1+3+3^2+\cdots+3^{m-1}=\frac{3^m-1}{2},$$

while $N_m<m!\,e$ by the Hilfssatz. The paper adds that this lower bound is of higher order than Dickson's bound $M=m^4-6m^3+13m^2-6m+1$ (p. 116) and exceeds it already for $m\ge7$.

Footnote 1 (p. 117). The paper states, without proof ("Es läßt sich noch zeigen"), that $N_m$ equals $(3^m-1)/2$ exactly only for $m\le3$.

Source. I. Schur, Über die Kongruenz $x^m+y^m\equiv z^m\pmod p$, Jahresber. Deutsch. Math.-Verein. 25 (1916), 114--117; the definition of $N_m$ and the start of the construction on printed p. 116, the rows, the example, the inequality $N_{m+1}\ge3N_m+1$, the bound $(3^m-1)/2$ and the footnote on printed p. 117, read on the page images. The copy read is identified on the source card.

Read depth. Claims checked: the definition, the construction, the inequality, the bound and the footnote were read clause by clause on the page images. The paper gives no proof of the construction beyond the example; the check sketched below is this page's, and nothing here is independently reviewed. The footnote's exactness claim is not checked.

Proof pointer

The paper leaves the construction to the reader (p. 117). A check written here: the last row holds the numbers $\equiv1\pmod3$, and the difference of two of them is $\equiv0\pmod3$. In a row built from an old row $R$, the differences of two entries are $3(x-y)$, $3(x-y)\pm1$ or $1$; those $\equiv1\pmod3$ cannot lie in the row, whose entries are $\equiv0,2\pmod3$, and $3(x-y)$ or $3(x-y)-1$ with $x>y$ in $R$ lies in the row only if $x-y$ lies in $R$, which the old distribution excludes. The rows cover $1,\ldots,3N_m+1$ because $3x$ and $3x-1$ for $x=1,\ldots,N_m$ fill the residues $0$ and $2$ up to $3N_m$. Induction from $N_1=1$ gives the bound.

Dependencies

The Hilfssatz of the same paper, for the finiteness of $N_m$ and the upper bound $N_m<m!\,e$ quoted beside it; the lower bound itself uses nothing else.

Bears on

• Problem 483: in the site's convention $f(m)=S(m)+1=N_m+1$, the bound gives $f(m)\ge(3^m+1)/2$, an exponential lower bound with base $3$; it does not touch the question whether $f(k)<c^k$, which asks for an upper bound.
• Problem 554: the site credits Schur with $C^k\ll R_k(K_3)$; the paper concerns integers only, and the passage to Ramsey numbers is the later difference coloring (color the edge $\{i,j\}$ of $K_{N_m+1}$ by the row of $|i-j|$), which gives $R_m(K_3)\ge N_m+2\ge(3^m+3)/2$, a translation the paper does not make.