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Odd numbers m >= 3 for which phi(2*m)/2 = phi(m)/2 is even, where phi = A000010 (Euler's totient).
5

%I #21 Dec 17 2024 08:24:31

%S 5,13,15,17,21,25,29,33,35,37,39,41,45,51,53,55,57,61,63,65,69,73,75,

%T 77,85,87,89,91,93,95,97,99,101,105,109,111,113,115,117,119,123,125,

%U 129,133,135,137,141,143,145,147,149,153,155,157,159,161,165,169,171,173,175,177,181,183,185,187,189,193

%N Odd numbers m >= 3 for which phi(2*m)/2 = phi(m)/2 is even, where phi = A000010 (Euler's totient).

%C This sequence is the complement of A197504 with respect to the positive odd numbers. It collects all positive odd numbers m with 4 dividing phi(m).

%C In the prime number factorization of an odd m >= 3 the largest even factor of phi(a(n)) is 2^{2*r1 + r3}, where r1 and r3 are the nonnegative number of distinct primes 1 (mod 4) and 3 (mod 4), respectively. This means that for m = a(n) one needs 2*r1 + r3 >= 2. See some examples below.

%C The number of solutions of the congruence x^2 == +1 (mod a(n)) or (inclusive) x^2 == -1 (mod a(n)) is 2^(r1 + r3) + delta_{r3,0}*2^r1, with 2*r1 + r3 >= 2, where r1 and r3 are the number of distinct primes 1 (mod 4) and 3 (mod 4), respectively, in the prime number factorization of a(n), and delta is the Kronecker symbol.

%C This follows from the result that primes 1 (mod 4) (A002144) have Legendre symbol (-1, p) = +1 and primes 3 (mod 4) (A002145) have (-1, p) = -1. The part (a) of the lifting theorem for powers of primes (Apostol, 5.30, pp. 121-122) is used. Also the theorem that for odd primes p there are exactly (p-1)/2 quadratic residues modulo p (and exactly (p-1)/2 nonresidues modulo p) is needed (see e.g., Silverman, ch. 20, Theorem 1, p. 151).

%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pp. 121-122.

%D J. H. Silverman, A Friendly Introduction to Number Theory, fourth ed., Pearson Education, Inc, 2014, ch. 20, pp. 149-155.

%H Michael De Vlieger, <a href="/A327922/b327922.txt">Table of n, a(n) for n = 1..10000</a>

%F All members of the set {odd m >= 1: 4 | phi(m)} ordered increasingly.

%e [n, a(n), [r1, r3], number of solutions x (mod a(n)), [solutions]] (with pm for + or -):

%e [1, 5, [1, 0], 4, [pm1, pm2]],

%e [5, 21 = 3*7, [0, 2], 4, [pm1, pm8]],

%e [20, 65 = 5*13, [1, 1], 8, [pm1, pm8, pm14, pm18]],

%e [34, 105 = 3*5*7, [1, 2], 8, [pm1, pm29, pm34, pm41]].

%t Select[Range[3, 200, 2], And[EvenQ[#1], #1 == #2] & @@ {EulerPhi[2 #]/2, EulerPhi[#]/2} &] (* _Michael De Vlieger_, Jun 28 2020 *)

%o (PARI) isok(m) = (m > 3) && (m % 2) && ((eulerphi(m) % 4) == 0); \\ _Michel Marcus_, Nov 13 2019

%Y Cf. A000010, A197504, A329584 (phi(a(n))/4).

%K nonn,easy

%O 1,1

%A _Wolfdieter Lang_, Nov 12 2019