Maths Olympiad Prep

Library / /6 of 14

Number theory Difficulty 5.9 AIME, harder Prove it Bulgaria

Find the least positive integer which divides 2n+152^n + 15 for some positive integer nn and has the form 3x24xy+3y23x^2 - 4xy + 3y^2 for some integers xx and yy.

Solution

Let d=3x24xy+3y2d = 3x^2 - 4xy + 3y^2 for some integers xx and yy and suppose that dd divides 2n+152^n + 15 for some positive integer nn. Obviously dd is odd and this implies that xx and yy have different parity. Then we have d3(mod4)d \equiv 3 \pmod{4}. Moreover, it follows from 3d=(3x2y)2+5y23d = (3x - 2y)^2 + 5y^2 that 3d(3x2y)2(mod5)3d \equiv (3x - 2y)^2 \pmod{5} and therefore 3d±1(mod5)3d \equiv \pm 1 \pmod{5} \Longleftrightarrow d±2(mod5)d \equiv \pm 2 \pmod{5} since (d,5)=1(d, 5) = 1.

Now d3(mod4)d \equiv 3 \pmod{4} and d±2(mod5)d \equiv \pm 2 \pmod{5} imply that d3(mod20)d \equiv 3 \pmod{20} or d7(mod20)d \equiv 7 \pmod{20}. It is obvious that d=3d = 3 is not a solution, and d=7d = 7 gives 2n1(mod7)2^n \equiv -1 \pmod{7}, which is also impossible. The next possibility d=23d = 23 satisfies the conditions of the problem for n=3n = 3 and x=2,y=1x = 2, y = -1. Therefore the required number is d=23d = 23.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.