MathLabs

Problem 3

Consider the following system of equations in the unknowns x1,x2,…,xnx_1, x_2, \ldots, x_n, where a,b,ca, b, c are real numbers with a≠0a \neq 0: ax12+bx1+c=x2,ax22+bx2+c=x3,…,axn−12+bxn−1+c=xn,axn2+bxn+c=x1.ax_1^2+bx_1+c=x_2,\quad ax_2^2+bx_2+c=x_3,\quad \ldots,\quad ax_{n-1}^2+bx_{n-1}+c=x_n,\quad ax_n^2+bx_n+c=x_1. Let Δ=(b−1)2−4ac\Delta=(b-1)^2-4ac. Prove that: (a) if Δ<0\Delta<0, the system has no solution; (b) if Δ=0\Delta=0, the system has exactly one solution; (c) if Δ>0\Delta>0, the system has more than one solution.
Step 1 of 5: Add all n equations to get a single sum condition
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In plain words

Adding a cyclic chain of equations is a classic trick: every xix_i appears once on the left (inside a quadratic) and once on the right (linearly), and the cyclic shift guarantees the right-hand sum is just a reordering of the same variables, so it can be cancelled against part of the left-hand side.

s(t):=at2+(b−1)t+c  ⟹  ∑i=1ns(xi)=∑i=1n(axi2+(b−1)xi+c)=0s(t):=at^2+(b-1)t+c \;\Longrightarrow\; \sum_{i=1}^n s(x_i)=\sum_{i=1}^n\bigl(ax_i^2+(b-1)x_i+c\bigr)=0
Detailed analysis

Sum all nn equations: ∑i=1n(axi2+bxi+c)=∑i=1nxi+1=∑i=1nxi\sum_{i=1}^n\bigl(ax_i^2+bx_i+c\bigr)=\sum_{i=1}^n x_{i+1}=\sum_{i=1}^n x_i. Moving the right-hand sum to the left gives ∑i=1n(axi2+(b−1)xi+c)=0\sum_{i=1}^n\bigl(ax_i^2+(b-1)x_i+c\bigr)=0. Defining s(t)=at2+(b−1)t+cs(t)=at^2+(b-1)t+c, this is exactly ∑i=1ns(xi)=0\sum_{i=1}^n s(x_i)=0.