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Szekeres and Turán found in 1937 a formula for the sum of the squares and the sum of the fourth powers of determinants of all $n$ by $n$ matrices with $\pm 1$ entries. (The sum of squares case follows easily from Cauchy-Binet identity.) Later Turán published a simpler proof for the sum of the fourth powers but in Chinese. I vaguely remember that there are simpler probabilistic proofs for both cases.

My question is about simple proofs for these identities, especially the one for 4th powers.

Is there a formula for the 6th power?

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    $\begingroup$ There is a proof of the fourth power result in the Chapter 5 exercises of Enumerative Combinatorics, vol. 2. I don't know how this compares with the argument of Szekeres and Turan. The proof technique completely breaks down for sixth powers. $\endgroup$
    – Richard Stanley
    Commented Jul 22, 2013 at 12:19
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    $\begingroup$ @RichardStanley: make that an answer. $\endgroup$
    – Zsbán Ambrus
    Commented Jul 22, 2013 at 15:21
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    $\begingroup$ Apart of combinatorial proof discussed above, one can prove the fourth moment formula by expanding the determinant wrt one row and then find recurrence relations $\endgroup$
    – Machinato
    Commented Aug 29, 2022 at 22:31

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This is an answer, but my insight is hardly a mathematical one. Nonetheless, here we go:

The 1955 paper of Turán is, in fact, published in both English and Chinese in the same journal. The Chinese version is pages 411-417 in the journal; the English language version is pages 417 - 423.

Turán writes: "I am pretty sure that on the way one can evaluate $M_n(6)$ or $S_n(6)$ after a little longer calculations and also further." But he doesn't give a formula.

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