Question

Consider the equations (n > 2)

$ax_1^2 + bx_1 + c = x_2$

$ax_2^2 + bx_2 + c = x_3$

$ax_3^2 + bx_3 + c = x_4$

...

$ax_{n-1}^2 + bx_{n-1} + c = x_n$

$ax_n^2 + bx_n + c = x_1$

in the variables $x_1, x_2, ...., x_n$

a, b, c be real numbers, a ≠ 0

Let us put $x_2 - x_1 = X_1, x_3 - x_2 = X_2, ......, x_n - x_{n-1} = X_{n-1}$ and $x_1 - x_n = X_n$

If $(b-1)^2 - 4ac = 0$, then the system has

• A. no solution
• B. n solutions
• C. (n-1) solutions
• D. one solution

Answer 1

Answer: (D)

one solution

Answer 2

Given the cyclic system

$ax_1^2 + bx_1 + c = x_2,$

$ax_2^2 + bx_2 + c = x_3,$

...

$ax_{n-1}^2 + bx_{n-1} + c = x_n,$

$ax_n^2 + bx_n + c = x_1,$

define the function

$f(x) = ax^2 + bx + c.$

Then the system can be written as

$f(x_i) = x_{i+1},$ with $x_{n+1}=x_1$.

A fixed point $\alpha$ of $f$ satisfies

$f(\alpha)=\alpha \implies a\alpha^2 + bx + c = \alpha,$

or

$a\alpha^2 + (b-1)\alpha + c = 0.$

The discriminant of this quadratic is given as

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

We are given $\Delta = 0$, so the quadratic has a double (unique) solution:

$\alpha = \frac{1-b}{2a}.$

Now, note that for any $x$,

$f(x)-x = a(x-\alpha)^2.$

Thus for each $i$:

$x_{i+1} - x_i = a (x_i-\alpha)^2.$

Summing these differences cyclically gives

$\sum_{i=1}^{n} (x_{i+1} - x_i) = a\sum_{i=1}^{n}(x_i-\alpha)^2.$

Since the left-hand side telescopes to 0, we have

$a\sum_{i=1}^{n}(x_i-\alpha)^2 = 0.$

Because $a\neq 0$ and squares are nonnegative, it follows that

$x_i - \alpha = 0$ for all $i$,

or

$x_1 = x_2 = \cdots = x_n = \alpha.$

Therefore, there is only one solution.