Indian Statistical Institute, ISI MMATH 2023 PMB Solutions & Discussions

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Problem 1.

Let $S=\left\{(x, y, z) \in \mathbb{R}^3 \mid x, y, z>0, x y+y z+z x=1\right\}$ . Prove that there exists $(a, b, c) \in S$ such that abc $\geq x y z$ for all $(x, y, z) \in S$ .

Topic: Set Theory, Inequalitites

Difficulty level: Easy

Solution:

Let $(x,y,z)\in S.$ By the AM-GM inequality

$(xyz)^{2/3}=\sqrt[3]{(xyz)^2}=\sqrt[3]{xyyzzx}\leq \frac{xy+yz+zx}{3}=\frac{1}{3}.$

Hence, for all $(x,y,z)\in S$

$xyz\leq \frac{1}{3^{3/2}}.$

Equality in the AM-GM inequality $\sqrt[n]{a_1\ldots a_n}\leq \frac{a_1+\ldots+a_n}{n}$ is achieved if and only if all numbers $a_1,\ldots,a_n$ are equal. Hence, to get the maximal possible product $xyz$ among all $(x,y,z)\in S$ we need to look for $(x,y,z)\in S$ with identical coordinates. If $x=y=z>0$ and $xy+yz+zx=1,$ then $3x^2=1$ and $x=\frac{1}{3^{1/2}}.$ Let $a=b=c=\frac{1}{3^{1/2}}.$ Since

$ab+bc+ca=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1,$

we have $(a,b,c)\in S.$ Further,

$abc=\frac{1}{3^{3/2}}\geq xyz$

for all $(x,y,z)\in S.$

Problem 2.

Prove that $x^n$ is a solution of the differential equation

$y^{\prime \prime}+P(x) y^{\prime}+Q(x) y=0$

on $[-1,1]$ for some nonzero real polynomials $P(x), Q(x)$ if and only if $n=1$ .

Topic: Differential Equations

Difficulty Level: Medium

Solution:

In this solution $n$ is assumed to be an integer.} Since $y=x^n$ must be twice differentiable on $[-1,1],$ we need to have $n\geq 0.$ If $n=0,$ then

$y''+P(x)y'+Q(x)y=Q(x).$

If $Q(x)=0$ for $x\in[-1,1]$ then $Q(x)=0$ for all $x\in \mathbb{R},$ which is not the case.

Let $n=1.$ If we take $y=x,$ $P(x)=x$ and $Q(x)=-1,$ then

$y''+P(x)y'+Q(x)y=x-x=0.$

So, $y=x$ is a solution to the equation of interest.

Assume that $y=x^n,$ $n\geq 2$ is a solution of

$y''+P(x)y'+Q(x)y=0, \ x\in [-1,1].$

Then

$n(n-1)x^{n-2}+nx^{n-1}P(x)+x^nQ(x)=0, \ x\in [-1,1].$

The function $R(x)=n(n-1)x^{n-2}+nx^{n-1}P(x)+x^nQ(x)$ is a polynomial that is equal to zero on the interval $[-1,1].$ It follows that all coefficients of $R$ are equal to zero. However, the coefficient of $R(x)$ near $x^{n-2}$ equals $n(n-1)\ne 0.$ The obtained contradiction shows that the only function $y=x^n$ which is a solution of the equation

$y''+P(x)y'+Q(x)y=0, \ x\in [-1,1],$

is $y=x$ (i.e. when $n=1$ ).

Problem 3.

Let $A=\left\{(x, y, z) \in \mathbb{R}^3 \mid x^2+y^2=1\right\}$ and $B=\mathbb{R}^2 \backslash\{(0,0)\}$ . Consider $A$ and $B$ as metric spaces via their inclusions in $\mathbb{R}^3$ and $\mathbb{R}^2$ respectively. Construct a bijection $f: A \rightarrow B$ such that both, $f$ and $f^{-1}$ are contimuous.

Topic: Metric Space

Difficulty Level: Medium

Solution:

Consider the mapping $f:A\to B$ defined by

$f(x,y,z)=(e^zx,e^zy).$

This mapping is well defined, as

$|f(x,y,z)|=e^z>0, \ f(x,y,z)\ne (0,0).$

The mapping $f$ is continuous, as functions $(x,y,z)\mapsto e^zx$ and $(x,y,z)\mapsto e^zy$ are continuous. The mapping $g:B\to A$ defined by

$g(u,v)=\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}},\ln \sqrt{u^2+v^2}\right)$

is also continuous (as $u^2+v^2>0$ for $(u,v)\in B$ ). In fact, $g=f^{-1}.$ Indeed,

$g(f(x,y,z))=g\left(e^zx,e^zy\right)=$

since $e^{2z}x^2+e^{2z}y^2=e^{2z}$

$=\left(\frac{e^zx}{\sqrt{e^{2z}}},\frac{e^zy}{\sqrt{e^{2z}}},\ln\sqrt{e^{2z}}\right)=(x,y,z),$

and

$f(g(u,v))=f\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}},\ln \sqrt{u^2+v^2}\right)=$

$=\left(e^{\ln \sqrt{u^2+v^2}}\frac{u}{\sqrt{u^2+v^2}}, e^{\ln \sqrt{u^2+v^2}}\frac{v}{\sqrt{u^2+v^2}}\right)=(u,v).$

So, the mapping $f$ is the needed mapping.

Problem 4.

Let $n \geq k \geq 2$ . A system consists of $n$ components, each of which functions independently with probability $1 / 2$ . The system is said to function if exactly $k$ of the $n$ components function. Compute the conditional probability that at least one of the first two components function given that the system functions.

Topic: Probability

Difficulty Level: Easy

Solution:

The system functions if and only if $k$ of the $n$ components function, and other $n-k$ components do not function. There are ${n\choose k}$ ways to choose $k$ components that function, hence the probability that the system functions equals ${n\choose k}2^{-n}.$ The system functions while first two components do not function if and only if there are exactly $k$ components among components $3,4,\ldots,n$ that function. So, the probability that the system functions and first two components do not function equals
${n-2\choose k}2^{-n}.$ It follows that the probability that first two components do not function conditionally on the event that the system functions equals

$\frac{{n-2\choose k}2^{-n}}{{n\choose k}2^{-n}}=\frac{{n-2\choose k}}{{n\choose k}}.$

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

$1-\frac{{n-2\choose k}}{{n\choose k}}.$

Problem 5.

Let $T: \mathbb{R}^d \rightarrow \mathbb{R}^d$ be a linear transfarmation where $d$ is a positive integer. Prove that $T^m\left(\mathbb{R}^d\right)=T^d\left(\mathbb{R}^d\right)$ for all $m \geq d$ .

Topic: Linear Algebra

Difficulty Level: Medium

Solution:

Denote by $T^0$ the identical operator. Since $T^{m+1}(\mathbb{R}^d)=T^m(T(\mathbb{R}^d))\subset T^m(\mathbb{R}^d),$ we have that

$T^0(\mathbb{R}^d)\supset T^1(\mathbb{R}^d)\supset \ldots \supset T^d(\mathbb{R}^d)\supset T^{d+1}(\mathbb{R}^d).$

Let $D_n$ denote the dimension of the subspace $T^n(\mathbb{R}^d).$ Then

$d=D_0\geq D_1\geq \ldots \geq D_d\geq D_{d+1}\geq 0.$

If all inequalities are strict, then

$D_1\leq d-1, D_2\leq d-2, \ldots, D_{d+1}\leq d-d-1\leq -1,$

which is impossible. So, there exists $n\in \{0,\ldots,d\}$ such that $D_n=D_{n+1},$ hence $T^n(\mathbb{R}^d)=T^{n+1}(\mathbb{R}^d).$ We will verify by induction that for all $k\geq 0$ $T^{n+k}(\mathbb{R}^d)=T^n(\mathbb{R}^d).$ For $k=0$ this is obvious. If $T^{n+k}(\mathbb{R}^d)=T^n(\mathbb{R}^d),$ then

$T^{n+k+1}(\mathbb{R}^d)=T(T^{n+k}(\mathbb{R}^d))=T(T^{n}(\mathbb{R}^d))=T^{n+1}(\mathbb{R}^d)=T^{n}(\mathbb{R}^d),$

where the latter equality follows from the choice of $n.$ So, for all $m\geq d(\geq n),$

$T^{m}(\mathbb{R}^d)=T^{n+m-n}(\mathbb{R}^d)=T^{n}(\mathbb{R}^d)=T^{n+d-n}(\mathbb{R}^d)=T^{d}(\mathbb{R}^d).$

Problem 6.

Let $B$ be a square matrix with values in $\{-1,+1,0\}$ such that every column has at most one +1 and at most one -1 . Show that $\operatorname{det}(B) \in\{-1,+1,0\}$ .

Topic: Linear Algebra

Difficulty Level: Medium

Solution:

Denote by $n$ the size of $B,$ i.e. $B$ is $n\times n$ matrix. We will prove the result by induction on $n.$ If $n=1,$ then

$B=\begin{bmatrix} b_{11}\end{bmatrix}$

and $\det B=b_{11}\in \{-1,0,+1\}.$ Assume the needed statement is true for $n-1$ and let

$B=\begin{bmatrix} b_{11} & b_{12} & \ldots & b_{1n} \\ b_{21} & b_{22} & \ldots & b_{2n} \\ \ldots & \ldots & \ldots & \ldots \\ b_{n1} & b_{n2} & \ldots & b_{nn} \end{bmatrix}.$

By assumption, each column has at most one $+1$ and at most one $-1.$ Consider following cases.

Case 1. There exists a column with $n-1$ zero elements. In other words, either all elements of the column are equal to zero, or there exists one element which is equal to $+1$ or to $-1.$ Then there are indices $k,l\in \{1,\ldots,n\}$ such that

$b_{kl}\in \{-1,0,+1\}, \ b_{il}=0, \ i\ne k.$

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

$\det B=\pm b_{kl} \det C,$

where the matrix $C$ is obtained by deleting $k$ -th row and $j-$ th column from the matrix $B$ . Every column of the matrix $C$ is contained in a column of the matrix $B.$ Hence, every column of the matrix $C$ contains at most one $+1$ and at most one $-1.$ By inductive assumption, $\det C\in \{-1,0,+1\}$ and $\det B\in \{-1,0,+1\}.$

Case 2. Every column of the matrix $B$ contains exactly one $+1$ and exactly one $-1.$ Consider the sum of all rows of the matrix $B.$ For each $j\in \{1,\ldots,n\}$

$\sum^n_{i=1}b_{ij}=0.$

Indeed, in the latter sum there is one $+1,$ one $-1,$ and $n-2$ zeros. Hence, rows of the matrix $B$ are linearly dependent and $\det B=0.$

In any case, $\det B\in \{-1,0,+1\}.$

Problem 7.

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

Topic: Group Theory

Difficulty Level: Easy

Solution:

Let $g\in G.$ Since $\langle a\rangle$ is a normal subgroup of $G,$

$g\langle a \rangle=\langle a \rangle g.$

In particular, $ga\in \{g,ag,a^2g\}.$

If $ga=g,$ then $a=e,$ which is not the case.

Assume that $ga=a^2g.$ The order of $g$ is an odd number. Denote it by $2n+1.$ Then

$a=g^{2n}a^2g.$

We will prove by induction that for $k=0,\ldots,n,$ $a=g^{2(n-k)}a^2g^{1+2k}.$ For $k=0$ this was already shown. Assume the result is proved for $k<n.$ Then

$\begin{aligned} a&=g^{2(n-k)}a^2g^{1+2k}=g^{2(n-k)-1}(ga)ag^{1+2k}=g^{2(n-k)-1}a^2(ga)g^{1+2k} \\ &=g^{2(n-k)-1}a^4g^{2+2k}=g^{2(n-k)-1}ag^{2+2k}=g^{2(n-(k+1))}(ga)g^{2+2k}\\ &=g^{2(n-(k+1))}a^2g^{3+2k}=g^{2(n-(k+1))}a^2g^{1+2(k+1)} \end{aligned}$

Hence, the result is true for all $k=0,\ldots,n.$ Taking $k=n$ we get

$a=a^2g^{1+2n}=a^2, \ a=e,$

which is impossible. So, $ga\ne a^2 g.$

Hence, $ga=ag$ and $a$ commutes with all elements of $G.$

Problem 8.

Let $p, q$ be odd primes such that $p-1$ divides $q-1$ . If $n$ is a positive integer coprime to $p q$ , prove that $p q$ divides $n^{q-1}-1$ .

Topic: Number Theory

Difficulty Level: Easy

Solution:

Remark. We need to assume that $p<q.$ If $p=q,$ then the statement is false. For example, we can take $p=q=3,$ $n=2.$ Then $n^{q-1}-1=2^2-1=3$ is not divisible by $pq=9.$ }

By Fermat’s little theorem $q$ divides $n^{q-1}-1.$ So, it is enough to prove that $p$ divides $n^{q-1}-1.$ Since $p-1$ divides $q-1,$ it follows that $q-1=k(p-1)$ for some integer $k\geq 2.$ Hence

$n^{q-1}-1=n^{k(p-1)}-1=(n^{p-1}-1)\left(n^{(k-1)(p-1)}+n^{(k-2)(p-1)}+\ldots +1\right).$

It follows that $n^{p-1}-1$ divides $n^{q-1}-1.$ By Fermat’s little theorem $p$ divides $n^{p-1}-1.$ Hence, $p$ divides $n^{q-1}-1$ and $pq$ divides $n^{q-1}-1.$