Mathematical Reflections 2013, Issue 1 - Problem U257

Problem:
a) Let $p$ and $q$ be distinct primes and let $G$ be a non-commutative group with $pq$ elements. Prove that the center of $G$ is trivial.

b) Let $p, q, r$ be pairwise distinct primes and let $G$ be a non-commutative group with
$pqr$ elements. Prove that the number of elements of the center of $G$ is either $1$ or a prime number.

Proposed by Mihai Piticari.

Solution:
We use the following

Lemma
If $G$ is a non-abelian group, then $G/Z(G)$ is not a cyclic group.

Proof
Homework!

a) Since $Z(G)$ is a subgroup of $G$, $|Z(G)|$ divides $|G|$, which means  $|Z(G)| \in \{1,p,q,pq\}$. Therefore, $|G/Z(G)| \in \{pq,q,p,1\}$ and since $G/Z(G)$ cannot be cyclic, it must be $|G/Z(G)|=pq$, i.e. $Z(G)$ is trivial.

b) As above, $|G/Z(G)|$ is a divisor of $G$, so, $|G/Z(G)| \in \{1,p,q,r,pq,qr,rp,pqr\}$. Since $G/Z(G)$ cannot be cyclic, the only possibilities are $|G/Z(G)| \in \{pq,qr,rp,pqr\}$, which means that $|Z(G)|$ is either $1$ or a prime number.

Mathematical Reflections 2013, Issue 1 - Problem U255

Problem:
Let $S_n$ be the group of permutations of $\{1,2,\ldots,n\}$. If $d > 1$ is an integer, let $H_d$ be the set of those $\sigma \in S_n$ for which there are $k \geq 1$ and $\sigma_1,\ldots,\sigma_k \in S_n$ with $\sigma = \sigma_1^d \cdots \sigma_k^d$. Find $H_2$ and $H_3$.

Proposed by Mihai Piticari and Sorin Radulescu.

Solution:
If $n=1,2$, clearly $H_2=H_3=S_n$. Let $n \geq 3$. We first prove that $H_d$ is a group for every integer $d > 1$. Indeed, if $\sigma, \tau \in H_d$, clearly $\sigma \tau \in H_d$. The associative property follows from the associativite property of the product of permutations. Moreover, $\textrm{id} \in S_n$ and $\textrm{id} = \textrm{id}^d$, so $\textrm{id} \in H_d$ for every $d > 1$. Finally if $\sigma_1,\ldots,\sigma_k \in H_d$ and $\sigma=\sigma_1^d \cdots \sigma_k^d$ for some $k \geq 1$, we have that $\sigma_1^{-1},\ldots,\sigma_k^{-1} \in S_n$ and $\tau=(\sigma_k^{-1})^d\cdots(\sigma_1^{-1})^d$ is the inverse of $\sigma$. Now, let $d=2$. It is clear that $H_2$ is a subgroup of $A_n$ since every permutation in $H_2$ is a product of even permutations and so is an even permutation. For every $a_1,a_2,a_3 \in \{1,\ldots,n\}$, we have also that $(a_1, a_2, a_3)=(a_1, a_3, a_2)^2$ so every $3$-cycle belongs to $H_2$. Since $A_n$ is generated by its $3$-cycles, it follows that $A_n$ is a subgroup of $H_2$, from which $H_2=A_n$. Now, let $d=3$. Obviously, $H_3$ is a subgroup of $S_n$. Moreover, for every $a_1,a_2 \in \{1,\ldots,n\}$ we have $(a_1,a_2)=(a_1,a_2)^3$, so every $2$-cycle belongs to $H_3$. Since $S_n$ is generated by its $2$-cycles, it follows that $S_n$ is a subgroup of $H_3$, which gives $H_3=S_n$.

Mathematical Reflections 2013, Issue 1 - Problem U253

Problem:
Evaluate $$\sum_{n>1} \dfrac{3n^2+1}{(n^3-n)^3}.$$

Proposed by Titu Andreescu.

Solution:
We observe that $$\dfrac{3n^2+1}{(n^3-n)^3}=\dfrac{1}{2}\left(\dfrac{1}{n^3(n-1)^3}-\dfrac{1}{(n+1)^3n^3}\right).$$
Hence, $$\begin{array}{lcl}\displaystyle \sum_{n=2}^\infty \dfrac{3n^2+1}{(n^3-n)^3}&=& \displaystyle \dfrac{1}{2} \sum_{n=2}^\infty \left(\dfrac{1}{n^3(n-1)^3}-\dfrac{1}{(n+1)^3n^3}\right)\\ &=& \displaystyle \dfrac{1}{2} \lim_{n \to \infty} \left(\dfrac{1}{8}-\dfrac{1}{(n+1)^3n^3}\right)\\&=& \dfrac{1}{16}. \end{array}$$

Mathematical Reflections 2013, Issue 1 - Problem S255

Problem:
Solve in real numbers the equation $$2^x+2^{-x}+3^x+3^{-x}+\left(\dfrac{2}{3}\right)^x+\left(\dfrac{2}{3}\right)^{-x}=9x^4-7x^2+6.$$

Proposed by Mihaly Bencze.

Solution:
We rewrite the equation in the form $$\left(2^{x/2}-2^{-x/2}\right)^2+\left(3^{x/2}-3^{-x/2}\right)^2+\left((2/3)^{x/2}-(2/3)^{-x/2}\right)^2=9x^4-7x^2.$$ Let $$f(x)=\left(2^{x/2}-2^{-x/2}\right)^2+\left(3^{x/2}-3^{-x/2}\right)^2+\left((2/3)^{x/2}-(2/3)^{-x/2}\right)^2$$ and $$g(x)=9x^4-7x^2.$$ Since $f(x)$ and $g(x)$ are even functions, it suffices to find the solutions when $x \geq 0$. It is easy to see that $x=0$ and $x=1$ are solutions of the given equation. Moreover, $f(x)$ is increasing for $x \geq 0$ since it is a sum of increasing functions, and $g(x)$ is increasing if $x \geq \sqrt{7/18}$ and decreasing if $0 \leq x \leq \sqrt{7/18}$, as can be seen from $g'(x)$. Since $f(x)$ and $g(x)$ are both injective functions if $x \geq 1$, then $h(x)=g(x)-f(x)$ is an injective function if $x \geq 1$, so $h(1)=0$ and $h(x) \neq 0$ for all $x>1$. Therefore the equation has no other solutions for $x \geq 0$, which means that the only solutions of the equation are $x=-1,0,1$.

Mathematical Reflections 2013, Issue 1 - Problem S253

Problem:
Solve in positive real numbers the system of equations:
$$\begin{array}{rcl}(2x)^{2013}+(2y)^{2013}+(2z)^{2013} & = & 3 \\ xy+yz+zx+2xyz & = & 1 \end{array} $$

Proposed by Roberto Bosch Cabrera.

Solution:
Let $2x=a, 2y=b, 2z=c$. Then, the given system of equations is equivalent to
$$\begin{array}{rcl}a^{2013}+b^{2013}+c^{2013} & = & 3 \\ ab+bc+ca+abc & = & 4. \end{array}$$ Using the AM-GM Inequality in the first equation, we get $abc \leq 1$ and from the second equation we get $ab+bc+ca \geq 3$. Suppose without loss of generality that $a \leq b \leq c$. By Chebyshev's Inequality, we have
\begin{equation}
1 \leq \dfrac{ab+bc+ca}{3} \leq \dfrac{(a+b+c)^2}{9},                           (1)
\end{equation}
therefore $a+b+c \geq 3$. Using Chebyshev's Inequality once again, we have $$\dfrac{a^{n-1}+b^{n-1}+c^{n-1}}{3} \leq \left(\dfrac{a+b+c}{3}\right)\left(\dfrac{a^{n-1}+b^{n-1}+c^{n-1}}{3}\right) \leq \dfrac{a^n+b^n+c^n}{3}$$ for all $n \in \mathbb{N}^*$. Since $\dfrac{a^{2013}+b^{2013}+c^{2013}}{3}=1$, we get $$a^n+b^n+c^n \leq 3$$ for all positive integers $n < 2013$, and in particular $a+b+c \leq 3$. This gives $a+b+c=3$, and from (1) we get $ab+bc+ca=3$, which implies $abc=1$. Therefore, $a+b+c=3\sqrt[3]{abc}$, so $a=b=c=1$, i.e. $x=y=z=1/2$ is the only solution to the given system of equations.

Mathematical Reflections 2013, Issue 1 - Problem J256

Problem:
Evaluate $$1^2 2!+2^2 3! + \ldots + n^2(n+1)!.$$

Proposed by Titu Andreescu.

Solution:
We have $$\begin{array}{lcl} k^2(k+1)! & = & [(k^2+k-2)-(k-2)](k+1)!\\ & = & [(k-1)(k+2)-(k-2)](k+1)!\\ & = &(k-1)(k+2)!-(k-2)(k+1)! \end{array}$$ for all $k \in \mathbb{N}$. Therefore, $$1^2 2!+2^2 3! + \ldots + n^2(n+1)!=(n-1)(n+2)!+2.$$

Mathematical Reflections 2013, Issue 1 - Problem J254

Problem:
Solve the following equation $F_{a_1} + F_{a_2} + \ldots + F_{a_k} = F_{a_1+a_2+\ldots+a_k}$ , where $F_i$ is the $i$-th
Fibonacci number.

Proposed by Roberto Bosch Cabrera.

Solution:
If $k=1$ the equation has infinitely many solutions. Let $k \geq 2$ and suppose that $1 \leq a_1 \leq a_2 \leq \ldots \leq a_k$.
We have five cases.

(i) $k=2$. Suppose that $a_1+a_2 \leq 4$. It's easy to see that $(1,2)$ and $(1,3)$ are solutions. Now, suppose that $a_1+a_2 \geq 5$. If $a_1=1$, then $a_2 \geq 4$ and $F_{a_1+a_2-2}>F_{a_1}$. If $a_1>1$, then $a_2>2$, so $F_{a_1+a_2-1}>F_{a_2}$ and $$F_{a_1+a_2}=F_{a_1+a_2-1}+F_{a_1+a_2-2} > F_{a_2}+F_{a_1},$$ i.e. the equation has no solution if $a_1+a_2 \geq 5$.

(ii) $k=3$. There are no solutions when $a_3=1$. If $a_3=2$, we have the solution $(1,1,2)$. It's easy to see that there are no more solutions such that $a_2=1$ and $a_3>2$. Suppose that $a_3 \geq 2$ and $a_2 \geq 2$. If $a_3=2$, we get $F_{a_1+a_2+a_3-1}>F_{a_3}$ and if $a_3>2$, we get $F_{a_1+a_2+a_3-2}>F_{a_1+a_2}$, so $$F_{a_1+a_2+a_3}=F_{a_1+a_2+a_3-1}+F_{a_1+a_2+a_3-2}>F_{a_3}+F_{a_1+a_2} \geq F_{a_3}+F_{a_2}+F_{a_1},$$ i.e. no solution if $a_3>2$.

(iii) $k=4$. A simple check shows that there are no solutions if $a_4=1$. If $a_4 \geq 2$ and $a_3=1$ there are no solutions, and if $a_3 \geq 2$, we can argue similarly to the previous case and conclude that $$F_{a_1+a_2+a_3+a_4} > F_{a_4}+F_{a_1+a_2+a_3} \geq F_{a_4}+F_{a_3}+F_{a_2}+F_{a_1},$$ i.e. no solution in this case.

(iv) $k=5$. If $a_5=1$, we immediately see that $(1,1,1,1,1)$ is a solution. If $a_5 \geq 2$ and $a_4=1$, there are no solutions and if $a_4 \geq 2$ we get $$F_{a_1+a_2+a_3+a_4+a_5}>F_{a_5}+F_{a_1+a_2+a_3+a_4}\geq F_{a_5}+F_{a_4}+F_{a_3}+F_{a_2}+F_{a_1}.$$

(v) $k>5$. Clearly, $a_1+a_2+\ldots+a_k \geq k$. If $a_k=1$, there are no solutions since $a_1+a_2+\ldots+a_k=k$ and $F_k>k$. If $a_k \geq 2$ and $a_{k-1}=1$, then $a_1+a_2+\ldots+a_{k-1}=k-1$ and
$$\begin{array}{lcl} F_{a_1+a_2+\ldots+a_k}&=&F_{a_1+a_2+\ldots+a_k-1}+F_{a_1+a_2+\ldots+a_k-2}\\&>&F_{a_k}+F_{a_1+a_2+\ldots+a_{k-1}} =  F_{a_k}+F_{k-1} \\ & \geq & F_{a_k}+k-1 \\ & = & F_{a_k}+F_{a_{k-1}}+\ldots+F_{a_1}. \end{array}$$
If $a_k \geq 2$ and $a_{k-1} \geq 2$, we have
$$\begin{array}{lcl} F_{a_1+a_2+\ldots+a_k} & > & F_{a_k}+ F_{a_1+a_2+\ldots+a_{k-1}}\\ & > & F_{a_k}+F_{a_{k-1}} + F_{a_1+a_2+\ldots+a_{k-2}} \\ & \geq & F_{a_k}+F_{a_{k-1}}+F_{a_{k-2}}+F_{a_1+a_2+\ldots+a_{k-3}} \\ & \vdots & \vdots \\ & \geq & F_{a_k}+F_{a_{k-1}}+\ldots+F_{a_2}+F_{a_1}-1 , \end{array}$$ which gives $F_{a_1+a_2+\ldots+a_k} > F_{a_k}+F_{a_{k-1}}+\ldots+F_{a_2}+F_{a_1}$, so there are no solutions in this case.

Mathematical Reflections 2013, Issue 1 - Problem J253

Problem:
Prove that if $a,b,c>0$ satisfy $abc=1$, then $$\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2} \leq \dfrac{3}{4}.$$

Proposed by Marcel Chirita.

Solution:
By the AM-HM Inequality, we have $$\dfrac{1}{ab+1+a+1} \leq \dfrac{1}{4}\left(\dfrac{1}{ab+1}+\dfrac{1}{a+1}\right)=\dfrac{1}{4}\left(\dfrac{c}{c+1}+\dfrac{1}{a+1}\right)$$ $$\dfrac{1}{bc+1+b+1} \leq \dfrac{1}{4}\left(\dfrac{1}{bc+1}+\dfrac{1}{b+1}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+1}+\dfrac{1}{b+1}\right)$$ $$\dfrac{1}{ca+1+c+1} \leq \dfrac{1}{4}\left(\dfrac{1}{ca+1}+\dfrac{1}{c+1}\right)=\dfrac{1}{4}\left(\dfrac{b}{b+1}+\dfrac{1}{c+1}\right).$$
Summing up the three inequalities, we obtain $$\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2} \leq \dfrac{1}{4} \left( \dfrac{a+1}{a+1}+\dfrac{b+1}{b+1}+\dfrac{c+1}{c+1}\right)=\dfrac{3}{4}.$$