Maths Olympiad Prep

Track / Stage 4 / 55 of 340 #315 of 1964

Problem 315

AMC 12 late, AIME early
Number theory Difficulty 4.6 Find the answer

Find all integers aa such that 5∣a3+3a+15 \mid a^{3}+3 a+1

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

We test all possible congruences of aa:

−a≡0:a3+3a+1≡1-a \equiv 0: a^{3}+3 a+1 \equiv 1

−a≡1:a3+3a+1≡1+3+1≡0-a \equiv 1: a^{3}+3 a+1 \equiv 1+3+1 \equiv 0

−a≡2:a3+3a+1≡8+6+1≡0-a \equiv 2: a^{3}+3 a+1 \equiv 8+6+1 \equiv 0

−a≡3:a3+3a+1≡27+9+1≡2−1+1≡2-a \equiv 3: a^{3}+3 a+1 \equiv 27+9+1 \equiv 2-1+1 \equiv 2

−a≡4:a3+3a+1≡−1−3+1≡2-a \equiv 4: a^{3}+3 a+1 \equiv-1-3+1 \equiv 2

Therefore, all numbers congruent to 1 and 2 are solutions.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.