Gazeta Matematica 6-7-8/2017, Problem 27406

Problem:
Prove that the number $$N=1+(1!)^2\cdot1\cdot3+\ldots+(n!)^2\cdot n\cdot(n+2)$$ is a perfect square for all $n \in \mathbb{N}^*$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that for all $k=1,2,\ldots,n$, we have $$\begin{array}{lll} (k!)^2\cdot k\cdot(k+2)&=&(k!)^2[(k^2+2k+1)-1]\\&=&(k!)^2[(k+1)^2-1]\\&=&[(k+1)!]^2-(k!)^2. \end{array}$$ Adding all these equalities, we get $$\begin{array}{lll} N&=&1+[(2!)^2-(1!)^2]+\ldots+[((n+1)!)^2-(n!)^2]\\&=&1+[(n+1)!]^2-(1!)^2\\&=&[(n+1)!]^2. \end{array}$$

Gazeta Matematica 6-7-8/2017, Problem 27391

Problem:
Find all triples of consecutive integers such that the sum of all the possible quotients between two numbers is an integer.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $n-1,n,n+1$ be three consecutive integers. We have to find all the integers $n$ such that $$\dfrac{n-1}{n}+\dfrac{n-1}{n+1}+\dfrac{n}{n-1}+\dfrac{n}{n+1}+\dfrac{n+1}{n-1}+\dfrac{n+1}{n}=N$$ is an integer. Observe that
$$\begin{eqnarray*} N&=&2+\dfrac{2n+1}{n-1}+\dfrac{2n-1}{n+1}\\&=&6+\dfrac{3}{n-1}-\dfrac{3}{n+1}\\&=&6+\dfrac{6}{n^2-1}. \end{eqnarray*}$$ Hence $N$ is an integer if and only if $6/(n^2-1)$ is an integer, i.e. if and only if $(n^2-1)|6$. Therefore, $n^2-1 \in \{\pm 1, \pm 2, \pm 3, \pm 6\}$. By an easy check we get $n=\pm 2$. We conclude that the only triples which satisfy the conditions are $(-3,-2,-1)$ and $(1,2,3)$.

Gazeta Matematica 3/2017, Problem 27353

Problem:
Solve in real numbers the equation $$\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}=\dfrac{x(x^2+3)}{3x^2+1}.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
 Let $f(x)=\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}$. An easy check shows that $f$ is a one-to-one correspondence on $\mathbb{R}$. Let us prove that $f^{-1}(x)=\dfrac{x(x^2+3)}{3x^2+1}$. Indeed, let us solve with respect to $x$ the equation $y=f(x)$. We have $$\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}} = y.$$ If $y=1$, we have $1-x=0$, i.e. $x=1$. Assume that $y \neq 1$. Set $u=\sqrt[3]{1+x}$ and $v=\sqrt[3]{1-x}$. Hence, $$\dfrac{u-v}{u+v}=y \implies u=\dfrac{v(y+1)}{1-y}.$$ Since $u^3+v^3=2$, then $$v^3\dfrac{(y+1)^3}{(1-y)^3}+v^3=2 \implies v^3=\dfrac{(1-y)^3}{3y^2+1}.$$ Therefore, $1-x=\dfrac{(1-y)^3}{3y^2+1}$, which gives $x=\dfrac{y(y^2+3)}{3y^2+1}$, so $f^{-1}(x)=\dfrac{x(x^2+3)}{3x^2+1}$. It follows that the real solutions to this equation are on the intersection between one of the two functions and the line $g(x)=x$. Let us take $f^{-1}(x)$ for simplicity. Then, it must be $$\dfrac{x(x^2+3)}{3x^2+1}=x.$$ Clearly, $x=0$ is a solution. If $x \neq 0$, then $x^2+3=1+3x^2$, which gives $x=\pm 1$. So, $x \in \{-1,0,1\}$.

Gazeta Matematica 2/2017, Problem 27335

Problem:
Find all real solutions $(x,y,z)$ with $x,y,z \in \left(\dfrac{3}{2},+\infty\right)$ to the equation
$$\dfrac{(x+1)^2}{y+z-1}+\dfrac{(y+2)^2}{z+x-2}+\dfrac{(z+3)^2}{x+y-3}=18.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
By the Cauchy-Schwarz Inequality, we have $$\left((y+z-1)+(z+x-2)+(x+y-3)\right)\cdot18 \geq \left((x+1)+(y+2)+(z+3)\right)^2,$$ i.e. $$(x+y+z+6)^2 \leq 36(x+y+z-3).$$ Let $t=x+y+z+6$. We have $$t^2 \leq 36(t-9),$$ i.e. $$(t-18)^2 \leq 0.$$ It follows that $t=18$, i.e.
$$\begin{equation}\label{first-eq} x+y+z=12. \end{equation}$$
The equality is attained when $$\dfrac{x+1}{y+z-1}=\dfrac{y+2}{z+x-2}=\dfrac{z+3}{y+z-3}=\lambda, \qquad \lambda \in \mathbb{R}.$$ So, $$\begin{array}{lll} \lambda(y+z)-x&=&\lambda+1 \\ \lambda(z+x)-y&=&2(\lambda+1) \\ \lambda(x+y)-z&=&3(\lambda+1). \end{array}$$ Adding side by side the three equations, we get $(2\lambda-1)(x+y+z)=6(\lambda+1)$ and by $\eqref{first-eq}$, we obtain $\lambda=1$. So,
$$\begin{array}{lll} y+z-x&=&2 \\ z+x-y&=&4 \\ x+y-z&=&6. \end{array}$$ Comparing these three equations with $\eqref{first-eq}$, we get $(x,y,z)=(5,4,3)$.

Mathematical Reflections 2017, Issue 6 - Problem U429

Problem:
Let $n \geq 2$ and let $A$ be a $n \times n$ matrix with positive real entries. Do there exist a $n \times n$ matrix $X$ with $(n-1)^2$ zero entries and a diagonal matrix $D$ such that $AX=D$?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Assume that there exist a $n \times n$ matrix $X$ with $(n-1)^2$ zero entries and a diagonal matrix $D$ such that $AX=D$.
Observe that $$\det(A)\det(X)=\det(D) \neq 0,$$ so in partcular $\det X \neq 0$.
Now, $X$ has $n^2-(n-1)^2=2n-1$ nonzero entries. Since $X$ has $n$ columns, then $X$ has at least one column with at most one nonzero entry. But $\det X \neq 0$, so $X$ has at least one column with exactly one nonzero entry, say the $k$-th column. Denote by $a_{ij}$, $x_{ij}$ and $d_{ij}$ the elements of $A$, $X$ and $D$, respectively. Then, $x_{hk} \neq 0$ for some $h$ and $x_{ik}=0$ for all $i \neq h$. Hence, for $m \neq k$ we have
$$0=d_{mk}=\sum_{i=1}^n a_{mi}x_{ik}=a_{mh}x_{hk},$$ contradiction. The conclusion follows.

Mathematical Reflections 2017, Issue 6 - Problem U427

Problem:
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a function defined by $f(x,y)=\mathbf{1}_{(0,1/y)}(x)\cdot\mathbf{1}_{(0,1)}(y)\cdot y$, where $\mathbf{1}$ is the characteristic function. Evaluate
$$\int_{\mathbb{R}^2} f(x,y) \ dx \ dy.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $f$ is clearly non-negative, by Tonelli's Theorem we can evaluate the double integral as iterated integrals, i.e.
$$\int_{\mathbb{R}^2} f(x,y) \ dx \ dy=\int_{\mathbb{R}} \left(\int_{\mathbb{R}} f(x,y) \ dy\right) \ dx.$$
Observe that
$$\renewcommand{\arraystretch}{2}\begin{array}{lll} \displaystyle \int_{\mathbb{R}} f(x,y) \ dy&=& \displaystyle \int_{\mathbb{R}} \mathbf{1}_{(0,1/y)}(x)\cdot\mathbf{1}_{(0,1)}(y)\cdot y \ dy\\&=& \displaystyle \int_{\mathbb{R}} \mathbf{1}_{(0,1/x)}(y)\cdot\mathbf{1}_{(0,1)}(y)\cdot y \ dy\\&=& \displaystyle \int_{\mathbb{R}} \mathbf{1}_{(0,\min(1,1/x))}(y)\cdot y \ dy. \end{array}$$
Since $$\mathbf{1}_{(0,\min(1,1/x))}(y)=\begin{cases} \mathbf{1}_{(0,1)}(y) & \textrm{if } 0 < x \leq 1 \\ \mathbf{1}_{(0,1/x)}(y) & \textrm{if } x > 1, \end{cases}$$
it follows that $$\int_{\mathbb{R}} \mathbf{1}_{(0,\min(1,1/x))}(y)\cdot y \ dy = \begin{cases} 1/2 & \textrm{if } 0 < x \leq 1 \\ 1/(2x^2) & \textrm{if } x > 1. \end{cases}.$$
So, $$\int_{\mathbb{R}^2} f(x,y) \ dx \ dy=\int_0^1 \dfrac{1}{2} \ dx+\int_{1}^{\infty} \dfrac{1}{2x^2} \ dx=1.$$

Mathematical Reflections 2017, Issue 5 - Problem O422

Problem:
Let $P(x)$ be a polynomial with integer coefficients having an integer root $k$ and $P(0) \neq 0$. Prove that if $p$ and $q$ are distinct odd primes such that $P(p)=p<2q-1$ and $P(q)=q<2p-1$, then $p$ and $q$ are twin primes.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Assume that $k \in \mathbb{Z}$ is a root of the polynomial with integer coefficients $P$. So, $P(k)=0$, which gives $P(x)=(x-k)Q(x)$ for some polynomial with integer coefficients $Q(x)$. Hence,
$$p=P(p)=(p-k)Q(p), \qquad q=P(q)=(q-k)Q(q).$$ It follows that $(p-k) \mid p$ and $(q-k) \mid q$, so $p-k \in \{\pm 1, -p\}$ and $q-k \in \{\pm 1, -q\}$. Since $p$ and $q$ are distinct primes, then $p-k \neq q-k$ and we have seven cases.

(i) $p-k=-1$ and $q-k=1$. It follows that $q-p=2$, i.e. $q=p+2$, so $p$ and $q$ are twin primes.

(ii) $p-k=-1$ and $q-k=-q$. It follows that $p=2q-1$, contradiction.

(iii) $p-k=1$ and $q-k=-1$. It follows that $p-q=2$, i.e. $p=q+2$, so $p$ and $q$ are twin primes.

(iv) $p-k=1$ and $q-k=-q$. It follows that $p=2q+1$, contradiction.

(v) $p-k=-p$ and $q-k=-1$. It follows that $q=2p-1$, contradiction.

(vi) $p-k=-p$ and $q-k=1$. It follows that $q=2p+1$, contradiction.

(vii) $p-k=-p$ and $q-k=-q$. It follows that $p-q=-p+q$, i.e. $p=q$, contradiction.

In conclusion, $p$ and $q$ are twin primes.

Mathematical Reflections 2017, Issue 5 - Problem U425

Problem:
Let $p$ be a prime number and let $G$ be a group of order $p^3$. Define $\Gamma(G)$ as the graph whose vertices represent the noncentral conjugacy class sizes of $G$ and two vertices are joined if and only if the two associated conjugacy class sizes are not coprime. Determine the structure of $\Gamma(G)$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
We have two cases.

(i) $G$ is abelian. In such a case, all the conjugacy classes are central, so $\Gamma(G)$ has no vertex, i.e. $\Gamma(G)$ is the null graph.

(ii) $G$ is nonabelian. Since $Z(G)$ is a subgroup of $G$, then $|Z(G)| \ | \ |G|$. Since $G$ is nonabelian, then $|Z(G)| \neq |G|$ and since $G$ is a $p$-group, then $|Z(G)| \neq 1$. It follows that $|Z(G)| \in \{p,p^2\}$. If $|Z(G)|=p^2$, then $|G/Z(G)|=p$, so $G/Z(G)$ is cyclic and this implies that $G$ is abelian, contradiction. So, it must be $|Z(G)|=p$. Now, take a noncentral conjuacy class $\textrm{Cl}(a)$. Clearly, $|\textrm{Cl}(a)| \neq 1$. By the Orbit-Stabilizer Theorem, we have that $$|\textrm{Cl}(a)|=[G:C_G(a)],$$ where $C_G(a)$ is the centralizer of $a$ in $G$. It follows that $|\textrm{Cl}(a)| \ | \ |G|$. Since $Z(G)=\bigcap_{a \in G} C_G(a)$ and $a \in C_G(a)$, but $a \notin Z(G)$, then $Z(G) \subset C_G(a)$, which gives $|Z(G)| < |C_G(a)|$. Since $C_G(a)$ is a subgroup of $G$, then $|C_G(a)| \ | \ |G|$, which implies that $|C_G(a)|=p^2$. So, $|\textrm{Cl}(a)|=p$, i.e. each noncentral conjugacy class has size $p$, so the graph is a complete graph. In order to determine the number of its vertices, we use the Conjugacy Class Equation $$|G|=|Z(G)|+\sum_{i}[G:C_G(x_i)]$$ and we get $$p^3=p+kp,$$ i.e. $k=p^2-1$. We conclude that $\Gamma(G)=K_{p^2-1}$.

Mathematical Reflections 2017, Issue 5 - Problem S423

Problem:
Let $p$ and $q$ be prime numbers such that $p^2+pq+q^2$ is a perfect square. Prove that $p^2-pq+q^2$ is prime.

Proposed by Alessandro Ventullo, Milan, Italy

First solution:
Let $p^2+pq+q^2=n^2$. Then, $(p+q)^2=n^2+pq$ where $n \in \mathbb{N}$, which gives $$(p+q-n)(p+q+n)=pq.$$ If $p=q$, then $(2p-n)(2p+n)=p^2$. Since $2p-n<2p+n$, then $2p-n=1$ and $2p+n=p^2$, i.e. $4p=1+p^2$, which gives no integer solution. If $p \neq q$, assuming without loss of generality that $p<q$, then since $p+q-n<p+q+n$, we have
$$\begin{array}{lll} p+q-n &=& 1 \\ p+q+n &=& pq, \end{array} \qquad \begin{array}{lll} p+q-n &=& p \\ p+q+n &=& q. \end{array}$$
The first system gives $2(p+q)=pq+1$, i.e. $(p-2)(q-2)=3$, which gives $p=3, q=5$. The second equation of the second system gives $p+n=0$, contradiction. So, we get $(p,q) \in \{(3,5),(5,3)\}$ and $p^2-pq+q^2=19$, which is prime.

Second solution:Observe that $p^2<p^2+pq+q^2<(p+q)^2$ for all primes $p,q$. Hence, $$p^2+pq+q^2=(p+k)^2,$$ where $k \in \{1,2,\ldots,q-1\}$, i.e. $$pq+q^2=2kp+k^2.$$ We get
$$p(2k-q)=(q-k)(q+k) \qquad (1)$$
Since the right-hand side is positive, it follows that $q<2k$. Moreover, since $p$ is prime, then $p \ | \ (q-k)$ or $p \ | \ (q+k)$.

(i) If $p \ | \ (q-k)$, then $q-k=ph$ for some $h \in \mathbb{N}$, so equation (1) becomes $$2k-q=h(k+q).$$ It follows that $(k+q) \ | \ (2k-q)$. Since $2k-q=2(k+q)-3q$, then $(k+q) \ | \ 3q$. If $q=3$, there is no solution. If $q \neq 3$, we have $k+q \in \{1,3,q,3q\}$. But $1<q<k+q<2q<3q$, so it must be $k+q=3$. Since $q$ is prime, this gives $q=2$ and $k=1$, but this would imply that $p \ | \ 1$, contradiction.

(ii) If $p \ | \ (q+k)$, then $q+k=ph$ for some $h \in \mathbb{N}$, so equation (1) becomes $$2k-q=h(q-k).$$ It follows that $(q-k) \ | \ (2k-q)$. Since $2k-q=2(k-q)+q$, then $(q-k) \ | \ q$. But $q-k<q$, so $q-k=1$. Equation $\eqref{first-eq}$ becomes $$p(q-2)=2q-1,$$ which implies $(q-2) \ | \ (2q-1)=[2(q-2)+3]$, i.e. $(q-2) \ | \ 3$. So, $q=3$ or $q=5$, which gives $p=5$ or $p=3$ respectively.

Therefore, $(p,q) \in \{(3,5),(5,3)\}$, which gives $p^2-pq+q^2=19$, which is prime.

Conjecture:
Let $k$ be a positive integer such that if $p^2+kpq+q^2$ is a perfect square, then $p^2-kpq+q^2$ is a prime. Then, $k=1$. Me and Robert Bosch tried to disprove this conjecture with Maple for small $k$, but it seems that $k$ is the only positive integer with this property.

Mathematical Reflections 2017, Issue 5 - Problem S422

Problem:
Solve in positive integers the equation $$u^2+v^2+x^2+y^2+z^2=uv+vx-xy+yz+zu+3.$$

Proposed by Proposed by Adrian Andreescu, Dallas, USA

Solution:
The given equation can be written as $$(u-v)^2+(v-x)^2+(x+y)^2+(y-z)^2+(z-u)^2=6.$$
Since $u,v,x,y,z$ are positive integers, then $x+y \geq 2$, which gives $(x+y)^2 \geq 4$. Since $(x+y)^2 \leq 6$, we conclude that $x+y=2$, so $x=y=1$ and
$$(u-v)^2+(v-1)^2+(1-z)^2+(z-u)^2=2.$$
So, exactly two of the summands on the LHS are equal to $1$ and the other are equal to $0$. We have six cases.

(i) $(u-v)^2=(v-1)^2=1$ and $(1-z)^2=(z-u)^2=0$. From the last equations we get $u=z=1$ and from the first equations we get $v=2$.

(ii) $(u-v)^2=(1-z)^2=1$ and $(v-1)^2=(z-u)^2=0$. From the last equations we get $v=1$ and $u=z$ and from the first equations we get $u=z=2$.

(iii) $(u-v)^2=(z-u)^2=1$ and $(v-1)^2=(1-z)^2=0$. From the last equations we get $v=z=1$ and from the first equations we get $u=2$.

(iv) $(v-1)^2=(1-z)^2=1$ and $(u-v)^2=(z-u)^2=0$. From the last equations we get $u=v=z$ and from the first equations we get $u=v=z=2$.

(v) $(v-1)^2=(z-u)^2=1$ and $(u-v)^2=(1-z)^2=0$. From the last equations we get $u=v$ and $z=1$ and from the first equations we get $u=v=2$.

(vi) $(1-z)^2=(z-u)^2=1$ and $(u-v)^2=(v-1)^2=0$. From the last equations we get $u=v=1$ and from the first equations we get $z=2$.

In conclusion, $$(u,v,x,y,z) \in \{(1,2,1,1,1),(2,1,1,1,2),(2,1,1,1,1),(2,2,1,1,2),(2,2,1,1,1),(1,1,1,1,2)\}.$$