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Assume that m is odd.
In this case det(M) is odd.
Furthermore, we assume that M is odd.
Moreover, if M is odd or ord p M is odd for an odd prime p, then.
Theorem 1.1 (i) For any integer M ≥ 2, we have (ii) Moreover, if M is odd or ord p M is odd for an odd prime p, then .
Case 1. Suppose that k = n − m is odd, where k ≥ 1.
Similar(28)
Let m be odd and let p be a prime dividing m.
But for m being odd, these polynomials in studies normalize condition in (2).
where n ′ = m 2 − 1 2. (e) Let m be odd and let p be a prime dividing m.
when m is even, where r = u 1 + 2 u 2 + ⋯ + ( m − 1 ) u m − 1. Proof Let m be odd.
If (n-m) is odd, then the inequality (2.4) holds.
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CEO of Professional Science Editing for Scientists @ prosciediting.com