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 5 of 5: Case Δ > 0: more than one solution
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In plain words

Two roots of the auxiliary quadratic hand us two independent "all-equal" solutions for free; the problem only asks for more than one solution, so there is no need to hunt for non-constant solutions at all — though in general more may exist.

Δ>0  ⟹  s(r1)=s(r2)=0, r1≠r2  ⟹  (r1,…,r1) and (r2,…,r2) both solve the system\Delta>0 \;\Longrightarrow\; s(r_1)=s(r_2)=0,\ r_1\neq r_2 \;\Longrightarrow\; (r_1,\ldots,r_1) \text{ and } (r_2,\ldots,r_2) \text{ both solve the system}
Detailed analysis

If Δ>0\Delta>0, the quadratic ss has two distinct real roots r1≠r2r_1\neq r_2, so s(r1)=s(r2)=0s(r_1)=s(r_2)=0, i.e. ar12+br1+c=r1ar_1^2+br_1+c=r_1 and ar22+br2+c=r2ar_2^2+br_2+c=r_2. Taking xi=r1x_i=r_1 for every ii satisfies every equation of the cyclic system, since each one reduces to ar12+br1+c=r1ar_1^2+br_1+c=r_1; likewise xi=r2x_i=r_2 for every ii also satisfies the whole system. This produces two genuinely different constant solutions (r1,…,r1)≠(r2,…,r2)(r_1,\ldots,r_1)\neq(r_2,\ldots,r_2), so the system has more than one solution.