Competition problem — Number theory

Library / /377 of 520

Number theory Difficulty 6.4 National olympiad Prove it

8. Prove: (i) The modulus 1 character is the principal character;
(ii) There is no primitive character modulo 2;
(iii) $\chi(n ; 4,0)$ (see equation (8)) is not a primitive character, $\chi(n ; 4,1)$ (see equation (9)) is a primitive character;
(iv) $\chi\left(n ; p^{a}, l\right)$ (see equation (28)) is a primitive character if and only if
$(l, p)=1 ;$
(v) $\chi\left(n ; 2^{a}, l_{-1}, l_{0}\right)(\alpha \geqslant 3$ , see equation (32)) is a primitive character if and only if
$2 \nmid l_{0} ;$
(vi) If $\chi(n ; k)$ satisfies equation (22) or equation (24), then $\chi(n ; k)$ is a (real) primitive character if and only if each character on the right-hand side of equation (22) or equation (24) is a (real) primitive character.

Solution

8. (i), (ii), (iii) direct verification; (iv), (v) use the corresponding expressions (28), (32), the previous problem (i), the relationship between the indices of a given primitive root $g$ (when $k=p^{\alpha}, p$ is an odd prime, for all $\left.\alpha \geqslant 1\right), n$ modulo $p^{\alpha_{1}}$ and modulo $p^{a_{2}}$ $\left(\alpha_{1}>\alpha_{2}\right)$ , and the relationship between the index sets of $-1,5, n$ modulo $2^{a_{1}}$ and modulo $2^{a_{2}}$ $\left(\alpha_{1}>\alpha_{2}\right)$ ; (vi) use the definition and the Chinese Remainder Theorem.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.