# Indian Statistical Institute, ISI MMATH 2023 PMB Solutions & Discussions

## Problem 1.

Let . Prove that there exists such that abc for all .

# Solution:

Let By the AM-GM inequality

Hence, for all

Equality in the AM-GM inequality is achieved if and only if all numbers are equal. Hence, to get the maximal possible product among all we need to look for with identical coordinates. If and then and Let Since

we have Further,

for all

## Problem 2.

Prove that is a solution of the differential equation

on for some nonzero real polynomials if and only if .

# Solution:

In this solution is assumed to be an integer.} Since must be twice differentiable on we need to have If then

If for then for all which is not the case.

Let If we take and then

So, is a solution to the equation of interest.

Assume that is a solution of

Then

The function is a polynomial that is equal to zero on the interval It follows that all coefficients of are equal to zero. However, the coefficient of near equals The obtained contradiction shows that the only function which is a solution of the equation

is (i.e. when ).

## Problem 3.

Let and . Consider and as metric spaces via their inclusions in and respectively. Construct a bijection such that both, and are contimuous.

# Solution:

Consider the mapping defined by

This mapping is well defined, as

The mapping is continuous, as functions and are continuous. The mapping defined by

is also continuous (as for ). In fact, Indeed,

since

and

So, the mapping is the needed mapping.

## Problem 4.

Let . A system consists of components, each of which functions independently with probability . The system is said to function if exactly of the components function. Compute the conditional probability that at least one of the first two components function given that the system functions.

# Solution:

The system functions if and only if of the components function, and other components do not function. There are ways to choose components that function, hence the probability that the system functions equals The system functions while first two components do not function if and only if there are exactly components among components that function. So, the probability that the system functions and first two components do not function equals
It follows that the probability that first two components do not function conditionally on the event that the system functions equals

Correspondigly, the probability that at least one of the first two components function given that the system functions equals

## Problem 5.

Let be a linear transfarmation where is a positive integer. Prove that for all .

# Solution:

Denote by the identical operator. Since we have that

Let denote the dimension of the subspace Then

If all inequalities are strict, then

which is impossible. So, there exists such that hence We will verify by induction that for all For this is obvious. If then

where the latter equality follows from the choice of So, for all

## Problem 6.

Let be a square matrix with values in such that every column has at most one +1 and at most one -1 . Show that .

# Solution:

Denote by the size of i.e. is matrix. We will prove the result by induction on If then

and Assume the needed statement is true for and let

By assumption, each column has at most one and at most one Consider following cases.

Case 1. There exists a column with zero elements. In other words, either all elements of the column are equal to zero, or there exists one element which is equal to or to Then there are indices such that

Expanding the determinant with respect to -th column we find that

where the matrix is obtained by deleting -th row and th column from the matrix . Every column of the matrix is contained in a column of the matrix Hence, every column of the matrix contains at most one and at most one By inductive assumption, and

Case 2. Every column of the matrix contains exactly one and exactly one Consider the sum of all rows of the matrix For each

Indeed, in the latter sum there is one one and zeros. Hence, rows of the matrix are linearly dependent and

In any case,

## Problem 7.

Let be a finite group having an odd number of elements. Suppose is an element of of order 3 such that the cyclic subgroup generated by is normal in . Prove that commutes with every element of .

# Solution:

Let Since is a normal subgroup of

In particular,

If then which is not the case.

Assume that The order of is an odd number. Denote it by Then

We will prove by induction that for For this was already shown. Assume the result is proved for Then

Hence, the result is true for all Taking we get

which is impossible. So,

Hence, and commutes with all elements of

## Problem 8.

Let be odd primes such that divides . If is a positive integer coprime to , prove that divides .

# Solution:

Remark. We need to assume that If then the statement is false. For example, we can take Then is not divisible by }

By Fermat’s little theorem divides So, it is enough to prove that divides Since divides it follows that for some integer Hence

It follows that divides By Fermat’s little theorem divides Hence, divides and divides