For a positive integer $n$, the equation $a^2 + b^2 + c^2 + n = abc$ 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 $$(2p-3q)^2-5q^2=-4$$(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.$(a,b,c) \rightarrow (b,c,bc-a)$