For a positive integer n, the equation a2+b2+c2+n=abc is given in the positive integers. Prove that: 1. There does not exist a solution (a,b,c) for n=2017. 2. For n=2016, a is divisible by 3 for all solutions (a,b,c). 3. There are infinitely many solutions (a,b,c) for n=2016.
Problem
Source: MEMO 2016 T8
Tags: number theory, number theory proposed, equation, modular arithmetic
ThE-dArK-lOrD
25.08.2016 21:02
Modulo 4. Modulo 3. Let (a,b,c)=(3x,3y,3) where x=15p,y=15q such that (2p−3q)2−5q2=−4(exist infinitely many by Pell's). Note that 2p-3q\equiv q\pmod{2}\implies p\in \mathbb{Z}.
rkm0959
26.08.2016 04:20
(1) even even even -> parity, even even odd -> mod 4, even odd odd -> parity, odd odd odd -> parity (2) mod 3 (3) Start with (3,45,45) and spam (a,b,c) \rightarrow (bc-a,b,c) while sorting
rterte
27.08.2016 09:09
What is MEMO?
rightways
27.08.2016 14:01
https://www.math.aau.at/MEMO2016/
quangminhltv99
29.08.2016 09:59
a) If all three are even, then 2\mid RHS and 2\not\mid LHS. If one is odd, the rest are even, then 4\mid RHS, and since 4\not\mid LHS, hence no root. The case with two odd one even and all odd can be solved by simply work with parity.
b) If none of a,b,c divisible by 3, then 3\mid LHS, while 3\not\mid RHS, hence no solution this case. In the other case, we see that all three are divisible by 3.
lazizbek42
11.01.2022 12:59
1.mod 4 2.mod 3 3.(a,b,c) \rightarrow (b,c,bc-a)