USAMO 1983 Problem 2 - Roots of Quintic

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Binomial-theorem

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Binomial-theorem

#1 h Aug 16, 2011, 9:50 PM • 4 Y

Y by ImSh95, Rounak_iitr, Adventure10, Mango247

Prove that the roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$ cannot all be real if $2a^2 < 5b$ .

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leshik

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leshik

#2 h Aug 16, 2011, 10:03 PM • 3 Y

Y by ImSh95, Adventure10, ehuseyinyigit

If given equation has five real roots, then the third derivative, which is quadratic polynomial has two real roots (just apply Rolle's theorem 3 times ). From the other hand, the inequality implies that its discriminant is negative.

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basketball9

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basketball9

#3 h Aug 17, 2011, 3:19 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

leshik wrote:

If given equation has five real roots, then the third derivative, which is quadratic polynomial has two real roots (just apply Rolle's theorem 3 times ). From the other hand, the inequality implies that its discriminant is negative.

Is there a non-calculus solution?

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AIME15

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AIME15

#4 h Aug 17, 2011, 4:34 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

The actual problem statement of USAMO 1983 #2 is:

Prove that the roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$ cannot all be real if $2a^2 < 5b$ .

No calculus required

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basketball9

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basketball9

#5 h Aug 19, 2011, 4:10 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Thanks. I just conjured up the solution with AM-GM.

Note to Binomial-Theorem- EDIT THE PROBLEM STATEMENT

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AopsUser101

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AopsUser101

#6 h Nov 26, 2019, 4:24 PM • 4 Y

Y by v4913, ImSh95, Adventure10, Mango247

EDIT: I made a mistake in my solution. Will fix later.

This post has been edited 1 time. Last edited by AopsUser101, May 2, 2020, 11:24 PM

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mynameisbob12

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mynameisbob12

#7 h May 2, 2020, 11:22 PM • 1 Y

Y by ImSh95

Wait does @AopsUser101's solution work? Because in the last two step, it says that $5(a^2-2b) \ge a^2$ , but how does that lead to $2a^2 \le 5b$ ?. And didnt you want to prove that $2a^2 \ge 5b$ ? Sorry, Im a bit confused

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AopsUser101

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AopsUser101

#8 h May 2, 2020, 11:24 PM • 5 Y

Y by v4913, ImSh95, Mango247, Mango247, Mango247

mynameisbob12 wrote:

Wait does @AopsUser101's solution work? Because in the last two step, it says that $5(a^2-2b) \ge a^2$ , but how does that lead to $2a^2 \le 5b$ ?. And didnt you want to prove that $2a^2 \ge 5b$ ? Sorry, Im a bit confused

Yes, there was a problem. I’ll fix is later, as time allows.

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peatlo17

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peatlo17

#9 h May 3, 2020, 2:23 AM • 1 Y

Y by ImSh95

My solution is pretty much the same as @AIME15, just wanted to elaborate on the AM-GM step.
Click to reveal hidden text

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mynameisbob12

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mynameisbob12

#10 h May 8, 2020, 3:57 AM • 1 Y

Y by ImSh95

Since $(r_1)^2+...+(r_5)^2 =a^2-2b$ , by Cauchy we know that $(a^2-2b)(5) \ge a^2$ ,so $5b \le 2a^2$ , so we have proved the contapositive, and so we are done

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Grizzy

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Grizzy

#11 h Dec 31, 2020, 4:20 AM • 2 Y

Y by ImSh95, Mango247

Let $r_1, r_2, r_3, r_4, r_5$ be the roots. Then by Vieta's, we have

$a = - \sum_{1 \le i \le 5}r_i$
and

$b = \sum_{1 \le i < j \le 5} r_ir_j.$
Then the inequality

$2a^2 < 5b$
becomes

$2\cdot\left(\sum_{1 \le i \le 5}r_i\right)^2 < 5 \cdot \left(\sum_{1 \le i < j \le 5} r_ir_j\right)$
or

$2\cdot\left(\sum_{1 \le i \le 5}r_i^2\right) < \sum_{1 \le i < j \le 5} r_ir_j.$
Now, assume for the sake of contradiction that none of the $r$ 's are unreal. Then the squares of the $r_i$ are nonnegative. This means, we can apply AM-GM to get that

$\frac{r_i^2 + r_j^2}{2} \ge r_ir_j$
for all $1 \le i < j \le 5$ . Summing this over all $i, j$ , we get

$2\cdot\left(\sum_{1 \le i \le 5}r_i^2\right) \ge \sum_{1 \le i < j \le 5} r_ir_j$
which is a contradiction, as desired. $\square$

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Maths_1729

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Maths_1729

#12 h Jan 1, 2021, 11:38 AM • 2 Y

Y by ImSh95, Mango247

Solution

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franzliszt

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franzliszt

#13 h Aug 19, 2021, 1:29 AM • 2 Y

Y by Tafi_ak, ImSh95

Funny problem.

Rewrite the quintic as $a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0.$ We will prove the contrapositive: if all roots to the quintic are real, then $2(a_4)^2\ge 5(a_3)$ .

Let the roots be $a,b,c,d,e$ and define the sums $P_k=a^k+b^k+c^k+d^k+e^k$ . By Newton, $\begin{align*}0&=a_5P_1+a_4\\&= a_5P_2+a_4P_1+a_3\cdot 2,\end{align*}$ but since $a_5=1$ , we have $P_1=-a_4$ and $P_2=-a_4P_1-2a_3 = (a_4)^2-2a_3$ . Therefore, $(a_4)^2=P_2+2a_3$ and we wish to show $2(P_2+2a_3)\ge 5(a_3)\Rightarrow 2P_2\ge a_3.$ In expanded form, this is $2(a^2+b^2+c^2+d^2+e^2)\ge ab+ac+ad+ae+bc+bd+be+cd+ce+de$ which is immediate by Muirhead since $(2,0,0,0,0)\succ (1,1,0,0,0)$ .

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ajax31

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ajax31

#14 h Aug 23, 2022, 11:31 AM • 2 Y

Y by ImSh95, Mango247

Storage

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HamstPan38825

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HamstPan38825

#15 h Mar 15, 2023, 10:29 PM • 2 Y

Y by ImSh95, centslordm

Let $r_1, r_2, r_3, r_4, r_5$ be the roots; assume FTSOC they are real. The condition implies $2(r_1+r_2+r_3+r_4+r_5)^2 < 5\sum_{\mathrm{sym}} r_1r_2 \iff 2\sum_{i=1}^5 r_i^2 < \sum_{\mathrm{sym}} r_1r_2,$ which is obviously absurd.

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cinnamon_e

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cinnamon_e

#16 h Jun 27, 2023, 8:44 PM • 1 Y

Y by ImSh95

Denote the roots by $r_1$ , $r_2$ , \dots, $r_5$ . By Vieta's, we have
$r_1+r_2+\dots+r_5=-a\qquad\text{and} \qquad\sum_{i<j}r_ir_j=b.$ Note that
$(r_1+r_2+\dots+r_5)^2=a^2=r_1^2+r_2^2+\dots+r_5^2 +2\sum_{i<j}r_ir_j=(r_1^2+r_2^2+\dots+r_5^2)+2b.$ If $r_1^2+r_2^2+\dots+r_5^2<0$ , then the roots cannot all be real, so if $a^2<2b$ , then the roots cannot be all real. $2a^2<5b$ is weaker than $a^2<2b$ , which implies the result.

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megarnie

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megarnie

#17 h Jul 29, 2023, 2:45 PM • 1 Y

Y by centslordm

Suppose that the roots were all real. Call the roots $r_1, r_2, r_3, r_4, r_5$ .

Note that by Vieta's, $a^2 = \left(\sum_{i=1}^5 r_i\right)\text{ and } b = \left( \sum_{1\le i< j \le 5} r_i r_j \right)$ Using the fact that $a^2 + b^2 \ge 2ab$ for any reals $a,b$ (follows from the fact that $(a-b)^2 \ge 0$ ), we find that $\begin{align*} 4 \sum_{i=1}^5 r_i ^2 = \sum_{1\le i < j \le 5} (r_i^2 + r_j^2) \ge 2 \left( \sum_{1\le i< j \le 5} r_i r_j \right) \\ \implies 2\sum_{i=1}^5 r_i ^2 \ge \left( \sum_{1\le i< j \le 5} r_i r_j \right),\\ \end{align*}$ Now, we have $\begin{align*} 2a^2 = 2\left(\sum_{i=1}^5 r_i\right) \\ = 2 \sum_{i=1}^5 r_i ^2 + 4 \left( \sum_{1\le i< j \le 5} r_i r_j \right) \\ \ge 5 \left( \sum_{1\le i< j \le 5} r_i r_j \right) = 5b,\\ \end{align*}$ a contradiction. Therefore the roots cannot be all real.

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Rainmaker2627

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Rainmaker2627

#18 h Aug 27, 2023, 4:21 PM

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Let $x_1,x_2,\dots,x_5$ be the roots of $p(x)=x^5+ax^4+bx^3+cx^2+dx+e$ . We note that $p$ has
$\sum_{i=1}^5 x_i=-a\quad\text{and}\quad\sum_{\text{sym}} x_ix_j=b$ $\sum_{i=1}^5 x_i^2=a^2-2b$ So in fact $2a^2<5b$ is equivalent to $4\sum_{i=1}^5 x_i^2<-\left(\sum_{i=1}^5 x_i\right)^2$ which immeditately implies $p$ cannot have $5$ real roots.

This post has been edited 1 time. Last edited by Rainmaker2627, Aug 27, 2023, 4:21 PM

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peppapig_

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peppapig_

#19 h Sep 16, 2023, 2:55 PM

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FTSOC, assume that all roots are real. Let these roots be WLOG $r_1\leq r_2\leq \dots \leq r_5$ . Note that we have that
$a=r_1+r_2+r_3+r_4+r_5,$ and
$b=\sum r_xr_y,$ for all pairs of positive integers $(x,y)$ such that $1\leq x\leq y\leq 5$ . By Muirhead's [ $(2,0,0,0,0)$ majorizes $(1,1,0,0,0)$ ], we have that
$2(r_1^2+r_2^2+\dots+r_5^2)\geq b,$ and adding $4b$ to both sides gives us that
$2(r_1^2+r_2^2+\dots+r_5^2)+2(2b)\geq 5b \iff{} 2a^2\geq 5b,$ a contradiction, finishing the problem.

This post has been edited 2 times. Last edited by peppapig_, Sep 16, 2023, 2:58 PM
Reason: Muirhead's & (x,y) specification

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DouDragon

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DouDragon

#20 h Oct 2, 2023, 8:50 PM

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sol

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joshualiu315

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joshualiu315

#21 h Oct 24, 2023, 5:46 PM

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Assume that all of the roots are real. Denote them by $r_1, r_2, r_3, r_4, r_5$ . Then,

$a = -\sum_{1 \le i \le 5} r_i$ $b = \sum_{1 \le i < j \le 5} r_i r_j.$
Now, it suffices to prove

$2 \left ( \sum_{1 \le i \le 5} r_i \right )^2 \ge 5 \sum_{1 \le i < j \le 5} r_i r_j$
which is equivalent to

$2 \sum_{1 \le i \le 5} r_i^2 \ge \sum_{1 \le i < j \le 5} r_i r_j.$
This results from summing $\frac{r_i^2+r_j^2}{2} \ge r_ir_j$ over $1 \le i < j \le 5$ . $\square$

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bjump

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bjump

#22 h Jan 21, 2024, 3:02 PM

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Assume for the sake of contradiction all roots are real, denote them by $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$
Then
$2a^{2}=2(x_{1}+x_{2}+x_{3}+x_{4}+x_{5})^{2}$ $5b=5\sum_{\text{cyc}} x_{1}x_{2}$ If
$2a^{2} < 5b$ $2(x_{1}+x_{2}+x_{3}+x_{4}+x_{5})^{2}<5\sum_{\text{cyc}} x_{1}x_{2}$ $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} <\sum_{\text{cyc}} x_{1}x_{2} <5\sum_{\text{cyc}} |x_{1}x_{2}|$ But, by AM-GM
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq 5\sum_{\text{cyc}} |x_{1}x_{2}|$ A contradiction $\square$

This post has been edited 2 times. Last edited by bjump, Jan 21, 2024, 3:02 PM

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Shreyasharma

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Shreyasharma

#23 h Jan 21, 2024, 5:32 PM

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Assume that they are all real, say $r_i$ .

We easily find that the condition implies,
$\begin{align*} 2\left( \sum_i r_i \right)^2 < 5\sum_{i < j} r_ir_j \end{align*}$ This expands as,
$\begin{align*} 2 \sum_i r_i^2 &< \sum_{i < j} r_ir_j\\ \end{align*}$ However note $r_i^2 + r_j^2 > 2r_ir_j$ . Then over all pairs of $(i, j)$ we find,
$\begin{align*} 4 \sum_i r_i^2 &\geq 2 \sum_{i < j} r_ir_j\\ 2 \sum_i r_i^2 &\geq \sum_{i < j} r_i r_j \end{align*}$ contradiction. $\square$

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Math4Life7

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Math4Life7

#24 h Feb 21, 2024, 4:25 AM

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Let $p$ , $q$ , $r$ , $s$ , and $t$ be the roots of the given.
We can see from vieta's that $2a^2 = 2(p+q+r+s+t)^2$ . We can also see that $5b = 5\sum_{sym} pq$ . We want to prove that $2(p+q+r+s+t)^2 \geq 5\sum_{sym} pq$ . Crossing out terms we get $2\sum_{sym} p^2 \geq \sum_{sym} pq$ which follows from Muirhead. $\blacksquare$

This post has been edited 1 time. Last edited by Math4Life7, Feb 21, 2024, 4:25 AM

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RedFireTruck

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RedFireTruck

#25 h Mar 9, 2024, 5:26 PM

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We will show that if all the roots are real, then we must have $2a^2\ge 5b$ . Let the roots be real numbers $p,q,r,s,t$ . Then we want to show that $2(p+q+r+s+t)^2\ge 5(pq+pr+ps+pt+qr+qs+qt+rs+rt+st)$ or $2p^2+2q^2+2r^2+2s^2+2t^2\ge pq+pr+ps+pt+qr+qs+qt+rs+rt+st.$ By AM-GM, $\frac{p^2+q^2}2\ge pq$ , and the desired result follows.

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jeff10

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jeff10

#26 h Jun 25, 2024, 1:23 PM

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Quick Solution; No AM-GM

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megahertz13

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megahertz13

#27 h Nov 3, 2024, 2:11 PM

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We will use contradiction; assume that the roots are real. Call them $x_1, x_2, \dots, x_5$ . Note that $a = -x_1-x_2-\dots-x^5$ and $b = x_1x_2+x_1x_3+\dots+x_4x_5.$ Since $2a^2<5b$ , we know that $2(x_1+x_2+\dots+x_5)^2<5(x_1x_2+x_1x_3+\dots+x_4x_5)$ or $2(x_1^2+x_2^2+\dots+x_5^2+2(x_1x_2+x_1x_3+\dots+x_4x_5))<5(x_1x_2+x_1x_3+\dots+x_4x_5).$ Simplifying, we have $2(x_1^2+x_2^2+\dots+x_5^2) < x_1x_2+x_1x_3+\dots+x_4x_5.$ This follows by Muirhead.

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soryn

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soryn

#28 h Nov 4, 2024, 3:20 AM

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Instructively.....

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Marcus_Zhang

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Marcus_Zhang

#29 h Mar 18, 2025, 7:40 PM

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Set variables for roots.

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ehuseyinyigit

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ehuseyinyigit

#30 h Mar 18, 2025, 9:46 PM

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MACLAURIN. Since the problem is itself just Maclaurin's Inequality, here is a proof without expanding $2a^2$ .

General Case
Prove that the roots of
$x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n=0$ cannot all be real if $2a_1^2<\dfrac{4n}{n-1}a_2$ .

Proof: Let the roots be $r_1,r_2,\cdots,r_n$ . Then
$\sum_{i=1}^{n}{r_i}=-a_1 \qquad \text{and} \qquad \sum_{i<j}{r_ir_j}=a_2$ Use Maclaurin to obtain necessary condition for roots being positive reals as
$a_1^2=\left(\sum_{i=1}^{n}{r_i}\right)^2\overbrace{\geq}^{Maclaurin} \dfrac{2n}{n-1}\sum_{i<j}{r_ir_j}=\dfrac{2n}{n-1}a_2$ Thus, we should have
$2a_1^2\geq \dfrac{4n}{n-1}a_2$ if not, $2a_1^2<\dfrac{4n}{n-1}a_2$ implying the roots cannot all be positive as desired.

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ehuseyinyigit

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ehuseyinyigit

#31 h Mar 18, 2025, 10:33 PM

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And why no one just used Maclaurin ?

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