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 3 of 5: Case Δ < 0: no solution
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In plain words

This is a clean parity-of-sign argument: a sum made entirely of same-signed nonzero numbers can never collapse to zero, no matter how the xix_i are chosen.

Δ<0  ⟹  s(t)≠0 ∀t  ⟹  s(xi) all same sign  ⟹  ∑i=1ns(xi)≠0\Delta<0 \;\Longrightarrow\; s(t)\neq0 \ \forall t \;\Longrightarrow\; s(x_i) \text{ all same sign} \;\Longrightarrow\; \sum_{i=1}^n s(x_i)\neq0
Detailed analysis

If Δ<0\Delta<0, the quadratic ss has no real root, so s(t)s(t) never vanishes and keeps the sign of aa for every real tt. Every s(xi)s(x_i) has that same sign, so the sum ∑i=1ns(xi)\sum_{i=1}^n s(x_i) is nonzero. This contradicts ∑i=1ns(xi)=0\sum_{i=1}^n s(x_i)=0, so no real solution (x1,…,xn)(x_1,\ldots,x_n) exists.