# Brilliant Polynomial Roots Contest - Season 1

Welcome to the first ever Brilliant Polynomial Roots Contest. This is inspired by many other contests in Brilliant. The aim is to improve the skills of Brilliant users in olympiad problems that ask you to find some functions involving the roots of a polynomial by vieta's, newton sums or other methods. The rules are:

1. I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

2. A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

3. Please make a substantial comment.

4. Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 36 hours. Then, you must post the solution and you have the right to post a new problem.

5. If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

6. No restriction in techniques you can use! use of calculus and roots of unity is allowed. use of cyclotomic polynomials and möbius inversion is also allowed.

7. It is NOT compulsory to post original problems. But make sure it has not been posted on Brilliant.

8. Your question must have a polynomial. it can be like $f(x)=\dfrac{x^2-1}{x-1}$ or $f(x)=x+1$. Both are allowed as long as the simplified form is a polynomial.

SOLUTION TO PROBLEM n

proof here

Question as

PROBLEM n

EDIT: for external discussion go to the discussion board

Note by Aareyan Manzoor
3 years, 8 months ago

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PROBLEM 7

If $a, b, c$ are the roots of the equation

$x^3 - 2x^2 - 3x - 4 = 0$

Show that $\frac{a^{2016} - b^{2016}}{a - b} + \frac{b^{2016} - c^{2016}}{b - c} + \frac{c^{2016} - a^{2016}}{c - a}$ is an integer.

This problem has been solved by Aareyan Manzoor but after time was finished, so Dev Sharma can Post the next problem

- 3 years, 8 months ago

Simple induction man, it not only holds for 2016 but for all natural no. n in the powers.

- 3 years, 8 months ago

PROBLEM 1 given the roots of $y^8-y^7+y^6-1=0$ are $y_1,y_2,...,y_7,y_8$ find the value of $\dfrac{1}{y_1^9}+\dfrac{1}{y_2^9}+...+\dfrac{1}{y_7^9}+\dfrac{1}{y_8^9}$ This problem has been solved by dev sharma

- 3 years, 8 months ago

SOLUTION TO PROBLEM 1

$y^8 - y^7 + y^6 - 1 = 0$

Now we have to form a polynomial whose roots are $\frac{1}{y_{1}}, .....$

Now, let $x = \frac{1}{y}$

then $y = \frac{1}{x}$

Putting it in the above equation, we get

$x^8 + x^2 - x + 1 = 0$

$x^8 = x - x^2 - 1$

$x^9 = x^2 - x^3 - x ...(1)$

Putting $x = \frac{1}{y_{1}},..., \frac{1}{y_{8}}$ repeatedly in the equation and add them all and using bit Newton Sum ($S_{1} = S_{2} = S_{3} = 0$)

- 3 years, 8 months ago

Let $w_k$ denote the 2015th primitive roots of unity. Find

$\sum _{n=1}^{\phi \left(2015\right)}w_n\prod _{k=1}^{\phi \left(2015\right)}w_k$

This problem has been solved by Aareyan Manzoor

- 3 years, 8 months ago

SOLUTION TO PROBLEM 4 you forgot to format your question properly! the proof anyways:

claim: the product of all the primitive roots of unity are 1.

proof: we know that each primitive root of unity can be written as $e^{\dfrac{2k\pi i}{n}}$.for k coprime to n. when we multiply all of these we find: $exp(\dfrac{2\pi i}{n}\sum_{1≤x≤2015,gcd(x,2015)=1} x)$ the formula for the sum of all co-primes less then n is $\dfrac{n\phi(n)}{2}$ view this. put that there to get $exp(\phi(n)\pi i)$ $\phi(n)$ is always even apart from n=1,2. so this will always be 1 for n>2. at 2015 it is one then. the rest is easy i guess. we put 1 to find $\sum w_n=\mu(2015)=\boxed{-1}$

- 3 years, 8 months ago

PROBLEM 6

Let the roots of $p(x)=x^3-2x-2$ be $x_1,x_2,x_3$. Evaluate: $\sum_{i=1}^{3} \dfrac{x_i+1}{x_i^2+x_i+1}$

This problem has been solved by Dev Sharma

- 3 years, 8 months ago

SOLUTION TO PROBLEM 6

$x^3 - 2x - 2 = 0$

$x^3 - 1 = 2x + 1 ...(1)$

Now, lets see what we need to find,

$\frac{x + 1}{x^2 + x + 1}$

Now multiplying $x - 1$ in numerator and denominator.

$\frac{x^2-1}{x^3 -1}$

Now using (1)

$\frac{x^2 - 1}{2x + 1}$

Now, using vieta

$x_{1} + x_{2} + x_{3} = 0$

$x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1} = -2$

$x_{1}x_{2}x_{3} = 2$

Now putting $x = x_{1}, x_{2}, x_{3}$ in the expression we want to find out and simplifying it, we get

$\frac{x_{1}^2}{2x_{1} + 1} + \frac{x_{2}^2}{2x_{2} +1} + \frac{x_{3}^2}{2x_{3} + 1} -(\frac{1}{2x_{1} + 1} + \frac{1}{2x_{2} + 1} + \frac{1}{2x_{3} + 1})$

Now using those vieta relations, we can get answer $\frac{-1}{3}$

- 3 years, 8 months ago

We can simplify our last expression like this -

$\frac{x_{1}^2}{2x_{1} + 1} + \frac{x_{2}^2}{2x_{2} +1} + \frac{x_{3}^2}{2x_{3} + 1} -(\frac{1}{2x_{1} + 1} + \frac{1}{2x_{2} + 1} + \frac{1}{2x_{3} + 1})$

= $\frac{x_{1}^2(4x_{2}x_{3} + 2x_{2} + 2x_{3} + 1) + x_{2}^2(4x_{1}x_{3} + 2x_{1} + 2x_{3} + 1) + x_{3}^2(4x_{1}x_{2} + 2x_{1} + 2x_{2} + 1)}{(2x_{1} + 1)(2x_{2} + 1)(2x_{3} + 1)}$ - $\frac{(4x_{2}x_{3} + 2x_{2} + 2x_{3} + 1) + (4x_{1}x_{3} + 2x_{1} + 2x_{3} + 1) + (4x_{1}x_{2} + 2x_{1} + 2x_{2} + 1)}{(2x_{1} + 1)(2x_{2} + 1)(2x_{3} + 1)}$

Now, simplify it and get the answer!!

- 3 years, 8 months ago

Natural approach, there's another that is easier but it still good, well done. Now let's see your problem!

- 3 years, 8 months ago

PROBLEM 3 find $\sum \ln(1-\omega^3)$ where $\omega$ are 2015th roots of unity≠1.

This problem has been solved by Julian Poon

- 3 years, 8 months ago

PROBLEM 5 given the polynomial $x^4+x^3-2x+11=0$ a,b,c,d are its roots. evaluate $\sum_{cyc}\dfrac{a^3(a+1)}{a^4+11}$

This problem has been solved by Alan Enrique Ontiveros Salazar

- 3 years, 8 months ago

SOLUTION TO PROBLEM 5

Surely there's an easier way, but this is a new way:

$x^4=-x^3+2x-11 \\ x^5=x^3+2x^2-13x+11 \\ x^6=x^3-13x^2+13x-11$

So $\dfrac{x^3(x+1)}{x^4+11}=\dfrac{2x-11}{2x-x^3}$

Now we say that $2x-11=(2x-x^3)(Ax^3+Bx^2+Cx+D)$. Expanding and substituting the first three relations we get:

$2x-11=(-3A+B+C-D)x^3+(13A-2B+2C)x^2+(-9A+13B-2C+2D)x-11A-11B+11C$

Matching the coefficients and solving the system we get $A=\dfrac{2}{15}$, $B=\dfrac{4}{15}$, $C=-\dfrac{3}{5}$ and $D=-\dfrac{11}{15}$.

Hence, $\dfrac{2x-11}{2x-x^3}=\dfrac{2}{15}x^3+\dfrac{4}{15}x^2-\dfrac{3}{5}x-\dfrac{11}{15}$

So we want to find $\displaystyle \dfrac{2}{15} \sum x_i^3+\dfrac{4}{15}\sum x_i^2-\dfrac{3}{5} \sum x_i - \dfrac{11}{15}(4)$

Using Newton's Sums we get $S_1=-1,S_2=0,S_3=2,S_4=11$ and $P_1=S_1=-1$, $P_2=S_1P_1-2S_2=1$, $P_3=S_1P_2-S_2P_1+3S_3=5$.

So the sum is $\dfrac{2}{15}(5)+\dfrac{4}{15}(1)-\dfrac{3}{5}(-1)-\dfrac{11}{15}(4)=\boxed{-\dfrac{7}{5}}$.

- 3 years, 8 months ago

Problem 15

Prove that there doesn't exist polynomials with integer coefficients $f_k(x)$(k=1,2,3,4), such that $9x+4=f_1(x)^3+f_2(x)^3+f_3(x)^3+f_4(x)^3$

solved by Xuming Liang

- 3 years, 8 months ago

Solution: Consider the third root of unity $w\ne 1$. Since $w^2=-1-w$, we can express $f_i(w)=aw+b$ for some integers $a,b$. Thus $f_i(w)^3=(aw+b)^3=3ab(b-a)w+a^3+b^3-3a^2b$. Note that the coefficient of $w$ is always even, therefore the coefficient of $w$ on the LHS of the equality is even. However the same coefficient on the RHS is odd, thus these polynomials cannot exist.

- 3 years, 8 months ago

dont you think you need to prove thet $f_i$ is linear?

- 3 years, 8 months ago

Problem 16

Among the polynomials $P(x), Q(x), R(x)$ with real coefficients at least one has degree two and one has degree three. If $P(x)^2 + Q(x)^2 = R(x)^2$ prove that one of the polynomials of degree three has three real roots.

Solved by Khang Nguyen Thanh

- 3 years, 8 months ago

SOLUTION OF PROBLEM 16: This is a problem of All-Russian MO 2002.

Define $\deg(P)$ to mean the degree of $P(x)$, and likewise $\deg(Q)$ and $\deg(R)$.

Observe that $\max\{\deg(P), \deg(Q)\} = \deg(R)$.

This is because $\deg(R^2) = \deg(P^2 + Q^2) = \max \{\deg(P^2), \deg(Q^2)\}$ (leading coefficients of $P^2$ and $Q^2$ are positive and do not cancel), and $2\deg(R) = \deg(R^2) = \max \{\deg(P^2), \deg(Q^2)\} = 2 \max \{\deg(P), \deg(Q)\}$.

Thus one of $P, Q$ (without loss of generality, $Q$) must have degree 2, and both $P$ and $R$ must have degree 3.

Now we write $P^2 = (R + Q).(R-Q)$. We know that both factors have degree 3.

Since $P$ has either 1 or 3 real roots, $R+Q$ and $R-Q$ either both have one real root or three real roots accordingly.

Assume that $P, R+Q$ and $R-Q$ each have one real root; let the root of $P$ be $r_1$.

Since $r_1^2|P^2$, this means that $r_1$ is a root of both $R+Q$ and $R-Q$, and is therefore a root of both $R$ and $Q$.

Now let $P_0, Q_0$, and $R_0$ be the three polynomials $P, Q$, and $R$ with the common root divided out.

We have $P_0^2 +Q_0^2 = R_0^2$.

Then all three polynomials have real coefficients, $P_0$ and $R_0$ have degree 2, and $Q_0$ has degree 1.

We also know that $P_0$ has 0 real roots, $Q_0$ has 1, and $R_0$ has 0 or 2.

Since the latter case would imply that $R$ has 3 real roots, we instead assume that $R_0$ has 0 real roots.

Writing $Q_0^2 = (R_0 + P_0).(R_0 - P_0)$, we see that one of the factors must have degree 2 and the other degree 0.

If the first factor has degree 2, we may write $P_0 = ax^2 + bx + c, R_0 = ax^2 + bx + d$ for real $a, b, c, d$.

Then $Q_0 = (d - c)(2ax^2 + 2bx + c + d)$. Since $P_0$ and $R_0$ have no real roots, we know that $b^2 - 4ac < 0$ and $b^2 - 4ad <0$.

Adding the two inequalities and multiplying by 2, this means $4b^2 - 8a(c + d) < 0$.

But then $Q_0$ has no real roots, a contradiction.

Similarly, if $R_0 + P_0$ has degree 0, $R_0 - P_0$ has degree 2, we write $P_0 = ax^2 + bx + c, R_0 = -ax^2 - bx + d$.

Then $Q_0 = -(c + d)(2ax^2 + 2b + (c - d))$.

Again, $P_0$ and $R_0$ have no real roots, so $b^2 - 4ac < 0$ and $b^2 + 4ad < 0$.

Adding and multiplying by 2 gives $4b^2 - 8a(c - d) < 0$, which again implies that $Q_0$ has no real roots.

Thus our original assumption is false, and either $P$ or $R$ must be a third-degree polynomial with 3 real roots.

- 3 years, 8 months ago

Clearly, $R(x)$ has degree $3$ and $P(x)$ and $Q(x)$ have degree $2$ and $3$ (W.L.O.G.)

Also,

$R(x) = Q(x) \pm i P(x)$

From the above equation, it is clear that if $R(x)$ has a real root, it is also a root of $P(x)$ and $Q(x)$. So, $R(x)$ has at most $2$ real roots (counted with multiciplity). (Since $P(x)$ has degree $2$ and non real roots occur in conjugate pairs)

If $R(x)$ has $2$ real roots (one with multiplicity), then we are done. If not, then let $\alpha$ be the common real root to all three polynomials.

Let $\beta_{1}$ and $\beta_{2}$ be the other two roots of $Q(x)$,

Since,

$Q^2 (x) = (R(x) + P(x))(R(x) - P(x))$

and the two polynomials in the brackets are not identical, the set of roots must be $\alpha$, $\beta_{1}$, $\beta_{1}$ and $\alpha, \beta_{2}, \beta_{2}$. Again, since non real roots occur in conjugate pair, we must have $\beta_{1}, \beta_{2} \in \mathbb R$. Thus $Q(x)$ has three real roots.

- 3 years, 8 months ago

PROBLEM 20:

If $a,b,c$ are roots of the polynomial $x^3-3x^2+2x+3=0$ , find the value of $\dfrac{1}{a^3+b^3} + \dfrac{1}{b^3+c^3} + \dfrac{1}{a^3+c^3}$

###### Easy , because its created by me ;)

This problem is solved by Aareyan Manzoor

- 3 years, 8 months ago

By newtons sum $P_3=0$. so $\sum \dfrac{1}{a^3+b^3}=\sum \dfrac{1}{-c^3}$ pu x=1/x to have $x^3+\dfrac{2}{3} x^2-x+\dfrac{1}{3}=0$ $-\sum x^3= -((\sum(x))((\sum x)^2-3(\sum x_1x_2))+3\prod x)$ the result follows from vietas.

- 3 years, 8 months ago

Yo @Aareyan Manzoor damn you wrote the solution first ,

- 3 years, 8 months ago

PROBLEM 2

If $1, x_{1}, x_{2}, ....., x_{48}$ are 49th roots of unity, then find

$\displaystyle\sum_{i=1}^{48} \frac{1}{(1 - x_{i})^3}$ This problem has been solved by Aareyan Manzoor

- 3 years, 8 months ago

SOLUTION TO PROBLEM 2 the polynomial with roots $x_i$ is $x^{48}+x^{47}+...+x+1=\dfrac{x^{49}-1}{x-1}=0$ let $y=\dfrac{1}{1-x}$ or $x=\dfrac{y-1}{y}$.then we substitute: $(\dfrac{y-1}{y})^{49}-1=0,y\neq 0$ $y^{49}=(y-1)^{49}$ $y^{48}-24y^{47}+376y^{46}-4324y^{45}+.....+\dfrac{1}{49}=0$ using newtons sum $\begin{array}{c}a p_1&=&s_1&=&24\\ p_2&=&s_1p_1-2s_2&=&-176\\ p_3&=&s_1p_2-s_2p_1+3s_3&=&\boxed{-276}\end{array}$ we are done.

- 3 years, 8 months ago

PROBLEM 8

Let $p(x)$ be a polynomial with integer coefficients. Assume that $p(a) = p(b) = p(c) = -1$, where $a, b, c$ are three different integers. Prove that $p(x)$ has no integral zeroes.

This problem has been solved by Svatejas Shivakumar

- 3 years, 8 months ago

SOLUTION TO PROBLEM 8

Lemma: If $f(x)$ is a polynomial with integeral coefficients and $a$ is an integeral root of $f(x)$ and $m$ is any integer different from $a$, then $a-m$ divides $f(x)$.

Proof: On dividing $f(x)$ by $x-m$ we get $f(x)=(x-m)q(x)+f(m)$, where $q(x)$ is a polynomial with integral coefficients. For $x=a$, we get $f(a)=0=(a-m)q(a)+f(m)$ or $f(m)=-(a-m)q(a)$. Hence $a-m$ divides $f(m)$.

Suppose $d$ is an integeral root of $p(x)$, then by the lemma, $d-a,d-b,d-c$ divides $-1$ . But $-1$ has only $2$ factors namely $-1,1$. Hence at least two of $a,b,c$ are the same but $a,b,c$ are different. Hence, $p(x)$ has no integral zeroes.

- 3 years, 8 months ago

PROBLEM 9

The polynomial $ax^3+bx^2+cx+d$ has integral coefficients $a,b,c,d$ with $ad$ odd and $bc$ even. Show that at least one zero of the polynomial is irrational.

Nobody was able to solve this,so Svatejas Shivakumar can post the next problem

- 3 years, 8 months ago

SOLUTION OF PROBLEM 9

Let $x_i,i=1,2,3$ be the rational roots of the given.polynomial. Then $(ax)^3+b(ax)^2+ac(ax)+a^2d=0$.

Setting $y=ax$,we get $y^3+by^2+acy+a^2d=0$.

$y_i$ are the three rational zeros of the above equation,i.e. they must be integers. Also, since they are divisors of $a^2d$, they must be odd. Since $y_1+y_2+y_3=-b$ and $y_1y_2+y_2y_3+y_3y_1=ac$, both $b$ and $ac$ must be odd,i.e., $b$ and $c$ are odd hence $bc$ is odd. This contradicts the assumption that $bc$ is even. Hence at least one zeros of the polynomial is irrational.

- 3 years, 8 months ago

PROBLEM 10

The polynomial $P(x)=x^{n}+a_{1}x^{n-1}+\ldots+a_{n-1}x+1$ with nonnegative coefficients $a_{1},\ldots,a_{n-1}$ has $n$ real roots. Prove that $P(2) \ge 3^{n}$.

This problem has been solved by Aditya Agarwal

- 3 years, 8 months ago

Because of the given condition (non-negative coefficients), the roots would be non-positive.

So the polynomial can be expressed as $P(x)=(x+b_1)...(x+b_n)$ where b_i=-x_i\>0), and \(x_i are the roots.

Now, by the AM-GM inequality, we get, $2+b_i\geq3{b_i}^{\frac13}$

Now by Vieta's Formulas, we have, that the product of the roots is $1$.

So $P(2)=(2+b_1)...(2+b_n)\geq3^n$

Those who are saying that the product $b_1b_2...b_n=-1$, it won't be. Because $b_i>0$, and thus the product has to be greater than $-1$, so the product would definitely by $1$. (The confusion arised because of the substitution, $b_i=-x_i$)

- 3 years, 8 months ago

PROBLEM 11:

Prove: For any polynomial $g(x)$, $\text{deg}(g(x))>1$, another polynomial $k(x)$ can be substituted for $x$, in such a way that $g(k(x))$ can be expressed as a product of non-constant polynomials. (All the polynomials have integral coefficients)

Nobody posted the solution, Aditya Agarwal posted the solution

- 3 years, 8 months ago

Solution for Problem 11:

It is evident that, $g(a)-g(x)$ is divisible by $a-x$. Now lets us take $a$, such that $a-x$, is divisible by $g(x)$, for example, $a=k(x)=x+p(x)$. So, $g(k(x))$ is divisible by $g(x)$. Now because $\text{deg}(g(k(x)))$ is greater than $\text{deg}(g(x))$, the second factor is not constant.

- 3 years, 8 months ago

Please post the solution and the next problem as well since no one has solved it within the time limit.

- 3 years, 8 months ago

@Aditya Agarwal here is a problem!

Problem 14

Let $P(x)$ be a polynomial of degree $n>1$ with integer coefficients, and let $k$ be a positive integer. Consider the polynomial $Q(x) = P( P ( \ldots P(P(x)) \ldots ))$, where $P$ occurs $k$ times. Prove that there are at most $n$ integers $t$ such that $Q(t)=t$.

Dan Shwarz,Romania

solved by Xuming Liang

- 3 years, 8 months ago

This is the fifth problem of IMO 2006, I recognized it from a documentary about the USA IMO team. I will outline the basic idea of the proof:

The first idea shows why the case $k=2$ is important(sufficient):

If $Q(s)=s$ but $P(s)\ne s$, then $P(P(s))=s$. In other words, if $s$ is a fixed point of $Q$ but not of $P$, then it is a fixed point of $P(P(x))$. So it suffices to consider(count) the fixed points of $k=2$.

We now prove the case for $k=2$:

If all fixed points of $P(P(x))$ are fixed points of $P(x)$, then the result holds because $P$ has at most $n$ fixed points.

If not, then there exist $a\ne b$ such that $P(a)=b, P(b)=a$. The next observation is that all pairs $(a',b')$ that satisfy the previous equations($a',b'$ need not to be distinct) have the same sum, i.e. $a+b=a'+b'=c$ for some constant $c$. Thus all the numbers in these pairs(fixed points of $P(P(x))$) are the roots to the polynomial $P(x)+x-c$. Since this has the same degree as $P$($n>1$), we are done.

I know this isn't the full solution, so feel free to fill in the steps.

- 3 years, 8 months ago

This has totally stumped me! I tried proving it for k=2. But even when I did that, I was unsure of what to do?

- 3 years, 8 months ago

Keep trying though!
The $k=2$ is a very crucial step!

- 3 years, 8 months ago

Problem 17

Consider the sequence of polynomials $\{P_n(x)\}_{n=1,2,3,...}$ such that $P_n(2\cos x)=2^n\cos nx, \forall x\in\mathbb{R},\forall n\in\mathbb{N}^*$.

Prove that: $\large 1\le\dfrac{\sqrt[n]{P_n(x)}-2}{x-2}\le n, \forall x>2, \forall n\in\mathbb{N}^*$

- 3 years, 8 months ago

solution to Problem 17 $P_n(2x)=2^n\cos(n\arccos(x))=2^nT_n(x)$ chebyshev polynomials used here(of the first kind). we know:$T_2(x)=2x^2-1 ,T_3(x)=4x^3-3x,T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$ we see hat cases 2,3 are allways satisfied.so $2x-2≤2\sqrt[n+1]{T_{n+1}(2x)}-2≤(n+1)(2x-2)\\ 2x≤2\sqrt[n+1]{2xT_n(2x)-T_{n-1}(2x)}≤(n+1)(2x-2)+2\\ x^{n+1}≤4xP_n(2x)-4P_{n-1}(2x)≤((n+1)(2x-2)+2)^{n+1}$ we had $(2x)^n≤P_n(2x)≤(n(2x-2)+2)^n\\-((n-1)(2x-2)+2)^n≤-P_{n-1}(x)≤-(2x)^{n-1}$ multiplying and adding $4(2x)^{n+1}-4((n-1)(2x-2)+2)^n≤4xP_n(2x)-4P_{n-1}(2x)≤4x(n(2x-2)+2)^n-4(2x)^{n-1}$ now notice that $4(2x)^{n+1}-4((n-1)(2x-2)+2)^n≥x^{n+1}\\4x(n(x-2)+2)^n-4(2x)^{n-1}≤((n+1)(2x-2)+2)^{n+1}$ Hence proved by induction.

- 3 years, 8 months ago

PROBLEM 19

If $x_1,x_2,x_3$ are the roots of $P(x)=x^3-9x^2+14x-1$, find all the possible values of $\dfrac{x_1}{x_2}+\dfrac{x_2}{x_3}+\dfrac{x_3}{x_1}$.

This problem is solved by both , Xuming and Nihar.

- 3 years, 8 months ago

SOLUTION TO PROBLEM 19:

First I let $a=\dfrac{x_1}{x_2} , b= \dfrac{x_2}{x_3} , c= \dfrac{x_3}{x_1}$.

Now using $x_1x_2x_3=1$ (By Vieta's) , we can easily obtain that $ab+bc+ac=\dfrac{1}{a} + \dfrac{1}{b}+\dfrac{1}{c}$ .

Again by using $x_1x_2x_3=1$ we obtain :

$\dfrac{b}{a}+ \dfrac{c}{b}+ \dfrac{a}{c} = x_1^3+x_2^3+ x_3^3 \\ =(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_1x_3)+3x_1x_2x_3 \\ = 9^3-3(9)(14)+3 \\ =\boxed{354}$

Again by using $x_1x_2x_3=1$ we obtain : $\dfrac{a}{b}+ \dfrac{b}{c}+ \dfrac{c}{a}=(x_1x_2)^3+(x_2x_3)^3+(x_1x_3)^3$. Now we let $k=x_1x_2 \ , \ m=x_2x_3 \ , \ n=x_1x_3$ So we have:

$k^3+m^3+n^3=(k+m+n)^3-3(k+m+n)(km+mn+kn) + 3(k^2m^2n^2)$

Again by using $x_1x_2x_3=1$ we obtain : $km+mn+kn=x_1+x_2+x_3$ and $k^2m^2n^2=1$. Thus by using Vieta's again ,

$k^3+m^3+n^3=(14)^3-3(14)(9)+3 = \boxed{2370}$

We also have $(a+b+c)\left(\dfrac{1}{a} + \dfrac{1}{b}+\dfrac{1}{c}\right) =3 + \dfrac{a}{b}+ \dfrac{b}{c}+ \dfrac{c}{a}+ \dfrac{b}{a}+ \dfrac{c}{b}+ \dfrac{a}{c}$ that is :

$(a+b+c)(ab+bc+ac)=3 + 2370+354 = \boxed{2727}$

We also have $\left(\dfrac{a}{b}+ \dfrac{b}{c}+ \dfrac{c}{a}\right)\left( \dfrac{b}{a}+ \dfrac{c}{b}+ \dfrac{a}{c}\right) = a^3+b^3+c^3+a^3b^3+b^3c^3+a^3c^3+3$

Now we let $p=ab \ , \ q=bc \ , r = ac$ , and we have $p^3+q^3+r^3=(p+q+r)^3-3(p+q+r)(pq+qr+pr) +3p^2q^2r^2$ . Now using $abc=1$ , we have $pq+qr+pr=ab+bc+ac$ ,

$p^3+q^3+r^3=(p+q+r)^3-3(p+q+r)(pq+qr+pr) +3p^2q^2r^2 \\ = (ab+bc+ac)^3-3(ab+bc+ac)(a+b+c)+3$

and we also have

$a^3+b^3+c^3=(a+b+c)^3 - 3(a+b+c)(ab+bc+ac)+3$

$p^3+q^3+r^3+a^3+b^3+c^3 = (ab+bc+ac)^3+ (a+b+c)^3-6(a+b+c)(ab+bc+ac)+6 \\ \Rightarrow \left(\dfrac{a}{b}+ \dfrac{b}{c}+ \dfrac{c}{a}\right)\left( \dfrac{b}{a}+ \dfrac{c}{b}+ \dfrac{a}{c}\right) = (ab+bc+ac)^3+ (a+b+c)^3-6(2727)+6 \\ \Rightarrow (ab+bc+ac)^3+ (a+b+c)^3 = 855336$

Let $a+b+c=f \ , \ ab+bc+ac = g$ , we have $f^3+g^3=855336$ and $fg = 2727$ and substituting $g=\dfrac{2727}{f}$ in the former equation and solving the obtained quadratic , we get $f=a+b+c = 29 , 94$.

# HUSH!

- 3 years, 8 months ago

Well done! :D

- 3 years, 8 months ago

Nice problem ! @Alan Enrique Ontiveros Salazar

- 3 years, 8 months ago

The answers are indeed $94,29$. The key to finding the answer is constructing equations using symmetric sums. Note that our desired sum is equal to $\displaystyle \sum_{cyc}x_1^2x_3$, which we denote $A$. Let $B$ denote its symmetric counterpart: $\displaystyle \sum_{cyc} x_1^2x_2$.
Note that $A+B=\displaystyle \sum_{sym} x_1^2x_2=9\cdot 14-3=123$