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Statement

Dickson's theorem (as stated on p. 114). The congruence

$$(1)\qquad x^m+y^m\equiv z^m\pmod p$$

has a solution in three integers $x,y,z$ coprime to $p$ as soon as the prime $p$ exceeds a bound $M$ depending only on $m$. Footnote 1 (p. 114) records that Dickson stated the theorem only for prime $m$; Schur's statement and proof carry no such restriction.

Schur's bound (p. 115). The theorem holds with $M=m!\,e+1$, $e$ the base of the natural logarithms: "Der Dicksonsche Satz ist also richtig, wenn $M$ gleich $m!\,e+1$ gesetzt wird."

Comparison (p. 116). Dickson had shown by cyclotomy, for prime $m$ only, that $M=m^4-6m^3+13m^2-6m+1$ suffices; the paper remarks that so good a bound cannot be reached by its method alone, since the best bound the method gives is governed by the largest $N_m$ admitting a difference-free distribution of $1,\ldots,N_m$ into $m$ rows, and $N_m\ge(3^m-1)/2$ (lower bound, p. 117).

Source. I. Schur, Über die Kongruenz $x^m+y^m\equiv z^m\pmod p$, Jahresber. Deutsch. Math.-Verein. 25 (1916), 114--117; Dickson's theorem and footnote 1 on printed p. 114, the deduction on pp. 114--115 and its conclusion on p. 115, the comparison with Dickson's bound on p. 116, read on the page images. The copy read is identified on the source card.

Read depth. Claims checked: the statement, footnote 1, the deduction and the conclusion were read clause by clause on the page images; the deduction is elementary but not independently reviewed.

Proof pointer

Pages 114--115. If $m\mid p-1$, write $p-1=mq$, take a primitive root $g$ modulo $p$ and let $r_\nu$ be the least positive residue of $g^\nu$; the rows $r_\mu,r_{\mu+m},\ldots,r_{\mu+(q-1)m}$ for $\mu=0,\ldots,m-1$ distribute $1,\ldots,p-1$. When $p-1>m!\,e$ the Hilfssatz gives $\mu$ and indices $\alpha,\beta,\gamma$ with $r_{\mu+\gamma m}-r_{\mu+\beta m}=r_{\mu+\alpha m}$, and dividing the corresponding congruence by $g^\mu$ shows that $x=g^\alpha$, $y=g^\beta$, $z=g^\gamma$ solve (1). If $m\nmid p-1$, let $d=\gcd(m,p-1)$; then $p-1>m!\,e\ge d!\,e$ gives a solution of $x^d+y^d\equiv z^d\pmod p$ prime to $p$, and every $d$-th power residue modulo $p$ is also an $m$-th power residue, so (1) is solvable too.

Dependencies

The Hilfssatz of the same paper; the existence of primitive roots modulo a prime, and the fact that $d$-th and $m$-th power residues modulo $p$ coincide when $d=\gcd(m,p-1)$, both used as known.

Bears on

No Erdős problem page cites this result; the problems the paper bears on use the Hilfssatz and the lower bound of p. 117.