I recently came across a question regarding quadratic equations. I encountered this question in a maths Olympiad based book.

Find all the positive integers n such that the equation

$a_{n+1}$$x^2$ -$2$$x$$\sqrt{\sum_{i=1}^{n+1}a_{i}^2}$ +$\sum_{i=1}^na_{i}$=0

has real roots for every choice of real numbers $a_{1},a_{2}....a_{n+1}$

My initial approach was to take the discriminant positive... But that didn't help much. Might be possible, I lack proper information. I didn't understand what is so special about $a_{n+1}$. Could have been any coefficient.Can any inequality be applied? I also tried to apply the inequality that root mean square of$a_1,a_2,....a_{n}$ is greater than the mean $a_1,a_2,....a_{n}$... Please point me in the right direction. Any help would be appreciated.

  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, on this site we use MathJaX to format our maths. Here you can find a basic tutorial. $\endgroup$ – Nizar May 12 '17 at 7:34
  • 1
    $\begingroup$ I have tried to edit the question... Please have a look again $\endgroup$ – user440009 May 12 '17 at 12:32

Let $ \Delta $ be the discriminant of the equation $$ \Delta=4\sum_{i=1}^{n+1}a_{i}^{2}-4a_{n+1}\sum_{i=1}^{n}a_{i} $$ We must have that $ \Delta \geq 0 $ which implies that $$ \sum_{i=1}^{n+1}a_{i}^{2} \geq a_{n+1}\sum_{i=1}^{n}a_{i} $$ Rewrite the last inequality as $$ a_{n+1}^{2}-a_{n+1}\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}a_{i}^{2} \geq 0 $$ Since this must hold for any $ a_{n+1} \in \mathbb{R} $, we must have that the new discriminant $$\Delta '=\big(\sum_{i=1}^{n}a_{i}\big)^{2}-4\sum_{i=1}^{n}a_{i}^{2} \leq 0 $$ And the last inequality must hold for any real numbers $ a_{1}, \dots , a_{n} $. We show that only $ n \in \{ 1,2,3,4\} $ satisfy this property.

So let $ n \geq 5 $and consider the example $ a_{i}=i $ $ \forall i \in \overline{1,n} $ . Then in this case, we must have that $$\Delta '=\big(\sum_{i=1}^{n}i\big)^{2}-4\sum_{i=1}^{n}i^{2}=\frac{n^{2}(n+1)^{2}}{4}-4\frac{n(n+1)(2n+1)}{6} \leq 0 $$ Dividing the last inequality by $ \frac{n(n+1)}{12} $, we see that we must have $$ 3n(n+1) -8(2n+1) \le 0 $$

Computation then shows that $$ n \in \big[\frac{13-\sqrt{265}}{6},\frac{13+\sqrt{265}}{6}\big] $$

But $ \sqrt{265}<17 $ so $ n<5 $. Therefore the only remaining possibilities are $ n \in \{1,2,3,4\} $ and it is easily verified that all of them actually occur.

Indeed, for $ n=1 $, we must have that $ a_{1}^{2}-4a_{1}^{2} \leq 0 $ for any real number $ a_{1} $ which clearly holds.

For $ n=2$, we must have that $$ (a_{1}+a_{2})^{2}-4(a_{1}^{2}+a_{2}^{2})\leq 0 $$ which can be rewritten as $$ -(a_{1}-a_{2})^{2}-2(a_{1}^{2}+a_{2}^{2}) \leq 0 $$ which again holds true for any real numbers $ a_{1} $ and $ a_{2} $.

For $ n=3 $ we must have that $$(a_{1}+a_{2}+a_{3})^{2}-4(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})\leq 0 $$ but by Cauchy-Scwarz, we have that $$(a_{1}+a_{2}+a_{3})^{2} \leq 3(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}) $$ so clearly the inequality needed for $ n=3 $ holds as well.

Finally, for $n=4 $ what we need is exactly the Cauchy-Schwarz inequality $$(a_{1}+a_{2}+a_{3}+a_{4})^{2} \leq 4(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}) $$

We conclude that $ n \in \{1,2,3,4\} $.

  • $\begingroup$ You're welcome! I'm glad I could help. $\endgroup$ – Raizen May 12 '17 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy