Maths Olympiad Prep

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Number theory Difficulty 6.2 National olympiad Find the answer

3. (i) Find all prime numbers for which -3 is a quadratic residue;
(ii) Find all prime numbers for which ±3 is a quadratic residue;
(iii) Find all prime numbers for which ±3 is a quadratic non-residue;
(iv) Find all prime numbers for which 3 is a quadratic residue and -3 is a quadratic non-residue;
(v) Find all prime numbers for which 3 is a quadratic non-residue and -3 is a quadratic residue;
(vi) Find the prime factorization of (100)2−3(100)^{2}-3 and (150)2+3(150)^{2}+3.

A number or a short expression. Spacing and $ signs are ignored.

Solution

3. (i) p≡1( mod 6)p \equiv 1(\bmod 6). (ii) p≡1( mod 12)p \equiv 1(\bmod 12). (iii) p≡5( mod 12)p \equiv 5(\bmod 12); (iv) p≡−1p \equiv-1 ( mod 12);(v)p≡−5( mod 12).(vi)(100)2−3(\bmod 12) ;(\mathrm{v}) p \equiv-5(\bmod 12) .(\mathrm{vi})(100)^{2}-3 的素因数 p≡±1( mod 12)p \equiv \pm 1(\bmod 12). 1002−3=13⋅769;1502+3100^{2}-3=13 \cdot 769 ; 150^{2}+3 的素因数 p≡1( mod 6)p \equiv 1(\bmod 6) 及 p=3p=3.
1502+3=3⋅13⋅577150^{2}+3=3 \cdot 13 \cdot 577

3. (i) p≡1( mod 6)p \equiv 1(\bmod 6). (ii) p≡1( mod 12)p \equiv 1(\bmod 12). (iii) p≡5( mod 12)p \equiv 5(\bmod 12); (iv) p≡−1p \equiv-1 ( mod 12);(v)p≡−5( mod 12).(vi)(\bmod 12) ;(\mathrm{v}) p \equiv-5(\bmod 12) .(\mathrm{vi}) The prime factors of (100)2−3(100)^{2}-3 are p≡±1( mod 12)p \equiv \pm 1(\bmod 12). 1002−3=13⋅769;100^{2}-3=13 \cdot 769 ; The prime factors of 1502+3150^{2}+3 are p≡1( mod 6)p \equiv 1(\bmod 6) and p=3p=3.
1502+3=3⋅13⋅577150^{2}+3=3 \cdot 13 \cdot 577

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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.