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Statement

For an integer $m\geq1$, put $d=2m+1$. Then, for $C_m=(I+\frac12P)^T$, where $P$ is a cyclic permutation matrix,

$$\det C_m=1+2^{-d}.$$

Proof

In the Leibniz expansion of $\det(I+tP)$, a product is nonzero only if each row takes its diagonal entry or its entry of $tP$. Once one row takes its entry of $tP$, that entry's column is the diagonal column of the next row along the cycle, which must therefore take its entry of $tP$ too. So only the identity and the full cycle contribute. Consequently

$$\det(I+tP)=1+\operatorname{sgn}(P)t^d.$$

As $d$ is odd, the $d$-cycle is an even permutation, of sign $(-1)^{d-1}=1$. Set $t=1/2$; transposition does not change the determinant.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §2, Lemma 2.4, p. 3. The edition read is named on the source card. The determinant expansion is finite and uses no external theorem beyond the Leibniz formula.

Bears on. #1.