7
$\begingroup$

Find all real numbers $a_1, a_2, \cdots , a_n$ such that \begin{cases} \sqrt{a_1^3-a_2}=a_3-1\\ \sqrt{a_2^3-a_3}=a_4-1\\ \vdots\\ \sqrt{a_{n-1}^3-a_n}=a_1-1\\ \sqrt{a_n^3-a_1}=a_2-1 \end{cases}

I try to square both side and add them all together. This give me

$\sum\limits_{i=1}^{n} a_i^3-\sum\limits_{i=1}^{n} a_i^2+\sum\limits_{i=1}^{n} a_i-n=0$
$\sum\limits_{i=1}^{n}(a_i^3-a_i^2+a_i-1)=0 $
$\sum\limits_{i=1}^{n}(a_i-1)(a_i^2+1)=0 $
And I get an answer of $a_1 = a_2 = \cdots =a_n=1$ But I can neither find more nor prove that this is the only solution.

$\endgroup$
0

1 Answer 1

6
$\begingroup$

You've made a great start, and actually almost have it. Since the $a_i$ are real numbers, the LHS equations' square roots must all be non-negative real values. We then get from the RHS of your equations for all $1 \le i \le n$ that

$$a_i - 1 \ge 0 \;\;\to\;\; (a_i - 1)(a_i^2 + 1) \ge 0$$

This means

$$\sum_{i=1}^{n}(a_i - 1)(a_i^2 + 1) \ge 0$$

with it being equal to $0$ iff each summation term is $0$ so, since $a_i^2 + 1 \ge 1$, we must have $a_i - 1 = 0$, i.e.,

$$a_i = 1 \;\forall\; 1 \le i \le n$$

As such, since this also satisfies all of the constraint equations, this shows that you have found the only solution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .