Problem 3

Consider the following system of equations in the unknowns

x_1, x_2, \ldots, x_n

, where

a, b, c

are real numbers with

a \neq 0

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

\Delta=(b-1)^2-4ac

. Prove that: (a) if

\Delta<0

, the system has no solution; (b) if

\Delta=0

, the system has exactly one solution; (c) if

\Delta>0

, the system has more than one solution.

Step 4 of 5: Case Δ = 0: exactly one solution

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In plain words

A double root is the borderline case between "no sign change" and "a genuine sign change": the parabola only just touches zero, so the sum-of-same-sign argument from the previous case still applies almost everywhere, except at the single touching point, which pins down the unique constant solution exactly.

\Delta=0 \;\Longrightarrow\; s(t)=a(t-r)^2,\ s(t)\ \text{same sign as } a \text{ except } s(r)=0 \;\Longrightarrow\; x_1=\cdots=x_n=r

Detailed analysis

If $\Delta=0$ , the quadratic factors as $s(t)=a(t-r)^2$ for its unique root $r$ , so $s(t)$ has the same sign as $a$ for every $t\neq r$ and $s(r)=0$ . By Step 1, $\sum_{i=1}^n s(x_i)=0$ ; since every term $s(x_i)$ is either zero (when $x_i=r$ ) or has the fixed sign of $a$ (when $x_i\neq r$ ), the only way the sum can vanish is if every term is zero, i.e. $x_i=r$ for all $i$ . Conversely $(r,\ldots,r)$ solves the system, since $s(r)=0$ means $ar^2+br+c=r$ . So the system has exactly one solution, $(r,r,\ldots,r)$ .