Auxiliary polynomials are useful for solving complicated polynomial problems. This problem from ISI Entrance (and a Soviet Olympiad) is a good example.

# Tag: polynomial

# Understand the problem

Prove that , and determine the cases of equality.

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# Start with hints

Hence equality holds if and only if one of the three variables is and the other two are . QED

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# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

# Understand the problem

Show that must be equal to or .

*Note:* It is not required to show the existence of such numbers .

##### Source of the problem

Germany MO 2019, Problem 6

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# Start with hints

Now, observe one thing that this set of equations is symmetric in (x,y,z).

Observe that we are required to comment on (x+y+z).

Rewriting the equations as:

and then summing gives us that

Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ).

Let’s consider the case, where all of x,y,z is not 1.

Solving the first expression

then plugging this into the second two gives:

as z is not equal to 1.

Plugging the latter into the former and simplifying gives:

Now, observe that we already know z = 1 is a solution. This gives rise to

Hence we can claim that \( t^3 – 7t^2 + 7t + 7 = 0 \) has three solutions x,y,z.

Hence, \( t^3 – 7t^2 + 7t + 7 = (t-x)(t-y)(t-z)\).

Therefore, by Vieta’s formula, x+y+z = 7.

QED.

# Watch video

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

# Understand the problem

##### Source of the problem

##### Topic

##### Difficulty Level

##### Suggested Book

# Start with hints

The first observation is that the given polynomial is always > 0 as it is of the form \( (f(x))^2 + 13 \).

Hence, it has no real roots.

So, if the polynomial can be factorized, then obviously the factors will not be linear.

An odd degree polynomial has atleast one real root. Hence, none of the factors will be odd degree.

Let us approach the method of contradiction.

Assume that the polynomial can be factorized into n+1 factors with integer coefficients.

Now, we know that the factors must have atleast degree 2, in fact only even degrees i.e. 2, 4, 6, …

Now, let’s see if all the factors have a degree more than 2, then the degree of the whole polynomial must be \( \geq 4n + 4 \), but the degree of the polynomial is 4n.

So, two of the factors must be of degree 2.

Note that and for all , because they cannot have a real root.

Now, we have to make use of the idea that the coefficients are integers. So, some idea of divisibility must come in.

The given polynomial is .

Try to find out some values of x, for which P(x) is easily determined without n.

See, x = 1 and 6 works as \( x^2 – 7x + 6 = (x-1)(x-6) \).

Also, we know that f(1).g(1) | P(1) = 13. So, one of them must be 1.

Say, f(1) = 1 = 1 + a + b.

Also, f(6) = 36 + 6a – a = 36 + 5a.

Now, f(6) | P(6) = 13. So, f(6) = 36 + 5a = 1 or 13.

This gives rise to a single case of a = -7; where \( f(x) = x^2 -7x + 7\).

But f(x) has real roots implying that P(x) has real roots. Hence contradiction.

QED.

# Watch video

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

# Understand the problem

.

##### Source of the problem

Unknown

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Challenges and Thrills in Pre College Mathematics

# Start with hints

What is the first thing that you check in a Polynomial Identity?

Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same . But did you observe something fishy?

Do the Degree trick now… You see it right?

Yes, on the left it is and on the RHS it is .

So, there are two cases now… Figure them out!

Case 1: … i.e. P(x) is either a quadratic or a constant function.

Case 2: has coefficient zero till .

We will study case 1 now.

Case 1: … i.e. P(x) is either a quadratic or a constant function.

= where .

Now, expand using , it gives … Now find out all such polynomials satisfying this property.

For e.g. is a solution.

If P(x) is constant, prove that .

Assume a general form of P(x) = $latex $and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.

Now, we have already solved it for quadratic or less degree.

- Always
**Compare the Degree of Polynomials**in identities like this. It provides a lot of information. - Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.

- Find all polynomials .
- Find all polynomials \( P(cP(x)) – d P(P(x)) = P(x)^{2} \) depending on the values of c and d.

# Watch video

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

# Understand the problem

##### Source of the problem

##### Topic

##### Difficulty Level

##### Suggested Book

# Start with hints

# Watch video

# Connected Program at Cheenta

#### Math Olympiad Program

# Similar Problems

Here is a video post that discusses the roots of a polynomial problem from RMO 2017 problem 3. Watch, learn and enjoy the video.

## Some useful links:

TIFR 2013 Problem 38 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.

The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem Type:True/False?

Let (V) be the vector space of polynomials with real coefficients in variable (t) of degree ( \le 9). Let (D:V\to V) be the linear operator defined by (D(f):=\frac{df}{dt}). Then (0) is an eigenvalue of (D).

*Hint:*

If 0 were an eigenvalue, what would be its eigenvector?

**Discussion:**

There are several ways to do this. One possible way is to find out the matrix representation of (D) with respect to standard basis ( {1,t,t^2,…,t^n})( or any other basis) and observe that it is a (strictly) upper triangular matrix with all diagonal entries 0 and therefore the determinant of (D) is 0. This implies that D is not injective, so there is some nonzero vector to which when (D) is applied gives the zero vector. Therefore, (D) has 0 eigenvalue.

Another way to do this is by observing that (D(1)=0(1)), therefore 0 is an eigenvalue of (D) with 1 as an eigenvector.

## HELPDESK

**What is this topic:**Linear Algebra**What are some of the associated concept:**eigenvector,eigenvalue**Book Suggestions:**Linear Algebra done Right by Sheldon Axler

Try this beautiful problem from TOMATO Subjective Problem no. 173 based on the Sum of Polynomials.

* Problem* : Sum of polynomials

Let be polynomials in , each having all integer coefficients, such that . Assume that is not the zero polynomial. Show that and

**Solution*** *:

As are integer coefficient polynomials so gives integer values at integer points.

Now as is not zero polynomial

for some

Then or as =integer

or

But it is given that

This implies . This is only possible if

Hence the values of x for which is non-zero, are all zero. The values of x for which , we have implying each is zero.

Therefore .

Finally implies . Since hence it is 1.

(Proved)

*Chatushpathi*

**Topic:**Theory of Equations**Central idea:**Sum of square quantities zero implies each of the quantities is zero.**Course:**I.S.I. & C.M.I. Entrance Program- Video: Primes and Polynomials

**Cauchy Schwarz Problem: Let be a polynomial with non-negative coefficients.Prove that if for ,then the same inequality holds for each .**

**Discussion: Cauchy Schwarz’s Inequality:** Suppose for real numbers (\ a_{i},b_{i}), where (\ i\in{1,2,\dots,n}) we can say that $${\sum_{i=1}^{n}a_{i}^2}{\sum_{i=1}^{n}b_{i}^2}=\sum_{i=1}^{n}{a_{i}b_{i}}^2$$.

**Titu’s Lemma: **Let (\ a_{i},b_{i}\in{\mathbb{R}}) and let (\ a_{i},b_{i}>0) for (\ i\in{1,2,\dots,n})

$$\sum_{i=1}^{n}\frac{a_{i}^2}{b_{i}}\ge\frac{{\sum_{i=1}^{n}a_{i}}^2}{\sum_{i=1}^{n}b_{i}}$$

**Proof of Cauchy Schwarz’s Inequality: **We can write (\sum_{i=1}{n}a_{i}^2=\sum_{i=1}^{n}\frac{a_{i}^2 b_{i}^2}{b_{i}^2}\ge\frac{{\sum_{i=1}^{n}a_{i}b_{i}}^2}{\sum_{i=1}^{n}b_{i}^2}) **(using Titu’s lemma)**

(=>{\sum_{i=1}^{n}a_{i}^2}{\sum_{i=1}^{n}b_{i}^2}\ge {\sum_{i=1}^{n}{a_{i}b_{i}}^2)

**Solution:** Let ,then

now

(**Cauchy Schwarz’s Inequality**)

now it is given that for ,so

therefore,