Gazeta Matematica 2/2015, Problem S:L15.52

Problem:
Solve in natural numbers:
$$x!+112482225=y^2.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $112482225 \equiv 11 \pmod{13}$ and the perfect squares modulo $13$ are $0,1,3,4,9,10,12$. So, we have no solutions for $x \geq 13$. Hence, $x<13$. If $x=1$, we have $112482226$, which is divisible by $2$, but not by $4$, so it is not a perfect square.
If $x=2$, we have $112482227$, whose last digit is $7$, so it not a perfect square. If $x=3$, we have $112482231$, which has the last two digit odd, so it is not a perfect square. If $x=4$, we have $112482249$, which is divisible by $3$, but not by $9$, so it is not a perfect square. If $x \in \{5,6,7,8,9\}$, we obtain numbers which are divisible by $5$, but not by $25$, so they are not perfect squares. If $x=10$, we obtain $116111025=3^2\cdot5^2\cdot516049$, which is not a perfect square. If $x=11$, we have $152399025=12345^2$, so $y=12345$. If $x=12$, we have $591483825=3^2\cdot5^2\cdot2628817$, which is not a perfect square. In conclusion, the equation has a unique solution: $x=11,y=12345$.

Mathematical Reflections 2015, Issue 3 - Problem U340

Problem:
Define $$A_n=\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+\ldots+\dfrac{n}{n^2+n^2}.$$
Find $$\lim_{n \to \infty} n\left(n\left(\dfrac{\pi}{4}-A_n\right)-\dfrac{1}{4}\right).$$

Proposed by Yong Xi Wang, China

Solution:
We have $$nA_n=B_n,$$ where $$B_n=\dfrac{1}{1+\left(\frac{1}{n}\right)^2}+\ldots+\dfrac{1}{1+\left(\frac{n}{n}\right)^2}.$$
$B_n$ is the Riemann lower sum of $f(x)=\dfrac{1}{1+x^2}$ over $[0,1]$ with partition $$P=\left\{\left[0,\dfrac{1}{n}\right],\left[\dfrac{1}{n},\dfrac{2}{n}\right],\ldots,\left[\dfrac{n-1}{n},1\right]\right\}.$$
Therefore, $$\lim_{n \to \infty} B_n=\int_{0}^1 \dfrac{1}{1+x^2} dx=\left[\arctan x\right]_{0}^1=\dfrac{\pi}{4}.$$
It follows that $$\lim_{n \to \infty} n\left(n\left(\dfrac{\pi}{4}-A_n\right)-\dfrac{1}{4}\right)=\lim_{n \to \infty} n \cdot \lim_{n \to \infty} \left(\dfrac{n\pi}{4}-nA_n-\dfrac{1}{4}\right)=\infty.$$


Note. The result obtained is different from the official solution. See the official solution.

Mathematical Reflections 2015, Issue 3 - Problem U338

Problem:
Determine the number of pairs $(a,b)$ such that the equation $ax=b$ is solvable in the ring $(\mathbb{Z}_{2015},+,\cdot)$.

Proposed by Dorin Andrica, Cluj-Napoca, Romania

Solution:
If $b=0$, then the given equation is solvable for any $a \in \{0,1,\ldots,2014\}$. Hence, there are $2014$ pairs of the form $(a,0)$. Let $b \neq 0$. If $a$ is invertible in $(\mathbb{Z}_{2015},+,\cdot)$, then the equation $ax=b$ in the ring $(\mathbb{Z}_{2015},+,\cdot)$ is solvable and the solution is unique. The number of invertible elements in $(\mathbb{Z}_{2015},+,\cdot)$ is $\varphi(2015)=\varphi(5)\varphi(13)\varphi(31)=4\cdot12\cdot30=1440$. For any invertible element and for any $b \in \{1,2,\ldots,2014\}$ there is a unique solution. Hence, there are $1440\cdot2014=29001600$ pairs $(a,b)$ with $(a,2015)=1$ and $b \neq 0$. Now, let $(a,2015)>1$. Then, also $(b,2015)>1$. We have three cases.

(i) Let $b=5k$, where $k \in \{1,2,\ldots,402\}$. For any $k$, we have $a \in \{5,10,\ldots,5k\}$. So, there are $1+2+\ldots+402=201\cdot403=81003$ pairs $(a,b)$ with $(a,2015)>1$ in this case.

(ii) Let $b=13k$, where $k \in \{1,2,\ldots,154\}$. For any $k$, we have $a \in \{13,26,\ldots,13k\}$. So, there are $1+2+\ldots+154=77\cdot155=11935$ pairs $(a,b)$ with $(a,2015)>1$ in this case.

(iii) Let $b=31k$, where $k \in \{1,2,\ldots,64\}$. For any $k$, we have $a \in \{31,62,\ldots,31k\}$. So, there are $1+2+\ldots+64=32\cdot65=2080$ pairs $(a,b)$ with $(a,2015)>1$ in this case.

Since in each case we have counted twice the pairs for which $b=65k$ or $b=155k$ or $b=403k$, reasoning as before, we have that the total number of pairs when $(a,2015)>1$ is $81003+11935+2080-465-78-10=94465$. So, the total number of pairs $(a,b)$ for which the equation $ax=b$ is solvable in $(\mathbb{Z}_{2015},+,\cdot)$ is
$$2014+29001600+94465=29098079.$$

Note. The result obtained is different from the official solution. See the official solution.

Mathematical Reflections 2015, Issue 3 - Problem U337

Problem:
Let $n$ be a positive integer. Find all real polynomials $f$ and $g$ such that $$(x^2+x+1)^n f(x^2-x+1)=(x^2-x+1)^n g(x^2+x+1),$$ for all real numbers $x$.

Proposed by Marcel Chirita, Bucharest, Romania 

Solution:
Observe that $x^2-x+1=(1-x)^2-(1-x)+1$. Hence, by substitution $x \mapsto 1-x$, we get $$(x^2-3x+3)^n f(x^2-x+1)=(x^2-x+1)^n g(x^2-3x+3)$$ and dividing by the original equation, we have
\begin{equation}\label{first-eq}
\dfrac{g(x^2-3x+3)}{g(x^2+x+1)}=\left(\dfrac{x^2-3x+3}{x^2+x+1}\right)^n.             (1)
\end{equation}
Moreover, since $x^2+x+1=(-1-x)^2-(-1-x)+1$, by substitution $x \mapsto -1-x$, we get $$(x^2+x+1)^n f(x^2+3x+3)=(x^2+3x+3)^n g(x^2+x+1)$$ and dividing by the original equation, we have $$\dfrac{f(x^2+3x+3)}{f(x^2-x+1)}=\left(\dfrac{x^2+3x+3}{x^2-x+1}\right)^n.$$
By substitution $x \mapsto -x$, we obtain
\begin{equation}\label{second-eq}
\dfrac{f(x^2-3x+3)}{f(x^2+x+1)}=\left(\dfrac{x^2-3x+3}{x^2+x+1}\right)^n.            (2)
\end{equation}
Comparing (1) and (2), we obtain $$\dfrac{f(x^2-3x+3)}{g(x^2-3x+3)}=\dfrac{f(x^2+x+1)}{g(x^2+x+1)}$$ for all $x \in \mathbb{R}$. It follows that $\dfrac{f(x^2-3x+3)}{g(x^2-3x+3)}$ and $\dfrac{f(x^2+x+1)}{g(x^2+x+1)}$ have the same derivative for all $x \in \mathbb{R}$. So,
$$\begin{array}{lll} & & (2x-3)\dfrac{f'(x^2-3x+3)g(x^2-3x+3)-f(x^2-3x+3)g'(x^2-3x+3)}{(g(x^2-3x+3))^2}=\\&=&(2x+1)\dfrac{f'(x^2+x+1)g(x^2+x+1)-f(x^2+x+1)g'(x^2+x+1)}{(g(x^2+x+1))^2}.\end{array}$$ for all $x \in \mathbb{R}$. Since $(1-x)^2+(1-x)+1=x^2-3x+3$, by substition $x \mapsto 1-x$ into the right-hand side, we have $$(2x-3)\dfrac{f'(x^2-3x+3)g(x^2-3x+3)-f(x^2-3x+3)g'(x^2-3x+3)}{g^2(x^2-3x+3)}=0$$ for all $x \in \mathbb{R}$. It follows that $\dfrac{f(x^2-3x+3)}{g(x^2-3x+3)}$ is constant on $\mathbb{R}$, i.e. $f(x)=\lambda g(x)$ for some $\lambda \in \mathbb{R}$. Substituting into the original equation, we get $\lambda=1$.
Now, from equation (2), we get
$$\dfrac{f(x^2-3x+3)}{(x^2-3x+3)^n}=\dfrac{f(x^2+x+1)}{(x^2+x+1)^n}$$ for all $x \in \mathbb{R}$. It follows that $\dfrac{f(x^2-3x+3)}{(x^2-3x+3)^n}$ and $\dfrac{f(x^2+x+1)}{(x^2+x+1)^n}$ have the same derivative for all $x \in \mathbb{R}$. So,
$$\begin{array}{lll} & & (2x-3)\dfrac{f'(x^2-3x+3)(x^2-3x+3)-nf(x^2-3x+3)}{(x^2-3x+3)^{n+1}}=\\&=&(2x+1)\dfrac{f'(x^2+x+1)(x^2+x+1)-nf(x^2+x+1)}{(x^2+x+1)^{n+1}}.\end{array}$$ for all $x \in \mathbb{R}$. By substitution $x \mapsto 1-x$ into the right-hand side, we get once again that $$\dfrac{d}{dx} \dfrac{f(x^2-3x+3)}{(x^2-3x+3)^n}=0$$ for all $x \in \mathbb{R}$, so $f(x)=cx^n$ for some $c \in \mathbb{R}$. It follows that $f(x)=g(x)=cx^n$, where $c \in \mathbb{R}$.

Mathematical Reflections 2015, Issue 3 - Problem J342

Problem:
Solve in positive real numbers the system of equations $$\dfrac{x^3}{4}+y^2+\dfrac{1}{z}=\dfrac{y^3}{4}+z^2+\dfrac{1}{x}=\dfrac{z^3}{4}+x^2+\dfrac{1}{y}=2.$$

Proposed by Titu Andreescu, University of Texas at Dallas

Solution:
Adding the three equations, we get $$\left(\dfrac{x^3}{4}+x^2+\dfrac{1}{x}-2\right)+\left(\dfrac{y^3}{4}+y^2+\dfrac{1}{y}-2\right)+\left(\dfrac{z^3}{4}+z^2+\dfrac{1}{z}-2\right)=0.$$
Let $f(t)=\dfrac{t^3}{4}+t^2+\dfrac{1}{t}-2$. We have
$$f(t)=\dfrac{t^4+4t^3-8t+4}{4t}=\dfrac{(t^2+2t-2)^2}{4t} \geq 0$$ for all $t \in \mathbb{R}^+$ and $f(t)=0$ if and only if $t^2+2t-2=0$, i.e. $t=\sqrt{3}-1$. It follows that $f(x)+f(y)+f(z)=0$ if and only if $x=y=z=\sqrt{3}-1$.

Mathematical Reflections 2015, Issue 3 - Problem J339

Problem:
Solve in positive integers the equation $$\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1}=1.$$

Proposed by Titu Andreescu, University of Texas at Dallas

Solution:
By cyclic permutation, we can assume $x \leq y \leq z$. Then, $$\dfrac{x-1}{x+1} \leq \dfrac{y-1}{y+1} \leq \dfrac{z-1}{z+1}.$$ By the AM-GM Inequality, we have  $$1=\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1} \geq 3\sqrt[3]{\dfrac{x-1}{x+1}\cdot \dfrac{y-1}{y+1}\cdot \dfrac{z-1}{z+1}} \geq 3\cdot\dfrac{x-1}{x+1},$$ which gives $x \leq 2$. We have two cases.

(i) $x=1$. Hence, $$\dfrac{y-1}{z+1}+\dfrac{z-1}{2}=1,$$ i.e. $$2y=-z^2+2z+5.$$ Then $-z^2+2z+5=2y \geq 2$, which gives $z^2-2z-3 \leq 0$, i.e. $(z-3)(z+1) \leq 0$. So, $z \leq 3$. We get the unique solution $y=1, z=3$.

(ii) $x=2$. Hence, $$\dfrac{1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{3}=1.$$ Since $y \geq 2$, then $$1>\dfrac{2}{z+1}+\dfrac{z-1}{3}.$$ Clearly, $z \leq 3$. If $z=2$, we get $(x,y,z)=(2,2,2)$. If $z=3$, we get a contradiction.

In conclusion, $(x,y,z) \in \{(1,1,3),(1,3,1),(3,1,1),(2,2,2)\}$.

Mathematical Reflections 2015, Issue 3 - Problem J337

Problem:
Prove that for each integer $n \geq 0$, $16^n + 8^n + 4^{n+1} + 2^{n+1} + 4$ has two factors greater than $4^n$.

Proposed by Titu Andreescu, University of Texas at Dallas

Solution:
We have $$\begin{array}{lll} 16^n + 8^n + 4^{n+1} + 2^{n+1} + 4&=& (16^n+4^{n+1}+4)+2^n(4^n+2)\\&=&(4^n+2)^2+2^n(4^n+2)\\&=&(4^n+2)(4^n+2+2^n), \end{array}$$ and it's clear that each factor is greater than $4^n$.

Mathematical Reflections 2015, Issue 2 - Problem J331

Problem:
Determine all positive integers $n$ such that $$n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}+\dfrac{1}{n!}\right)$$ is divisible by $n$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Clearly, $n=1$ satisfies the condition. Let $n>1$. Then, $$n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}+\dfrac{1}{n!}\right)=n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n-1}\right)+(n-1)!+1.$$ It's easy to see that $$n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n-1}\right) \equiv 0 \pmod{n}.$$ Furthermore, by Wilson's Theorem, $(n-1)!+1 \equiv 0 \pmod{n}$ if and only if $n$ is prime, so $n$ must be a prime number.
In conclusion, $n=1$ or $n=p$, where $p$ is a prime number.

Mathematical Reflections 2015, Issue 1 - Problem U327

Problem:
Let $(a_n)_{n \geq 0}$ be a sequence of real numbers with $a_0=1$ and $$a_{n+1}=\dfrac{a_n}{n^2a_n+a^2_n+1}.$$
Find the limit $\displaystyle \lim_{n \to \infty} n^3 a_n$.

Proposed by Khakimboy Egamberganov.

Solution:
Observe that $a_{n+1}<\dfrac{a_n}{n^2a_n}=\dfrac{1}{n^2}$, so $\displaystyle \sum_{n=0}^{\infty} a_n$ converges by the Comparison Test. Let $$\sum_{n=0}^{\infty} a_n=c \in \mathbb{R}.$$
We have $$\dfrac{1}{a_{n+1}}=n^2+a_n+\dfrac{1}{a_n}.$$ So, $$\begin{array}{rcl} \dfrac{1}{a_1}&=&0^2+a_0+\dfrac{1}{a_0} \\ \dfrac{1}{a_2}&=&1^2+a_1+\dfrac{1}{a_1} \\ \vdots & \vdots & \vdots \\ \dfrac{1}{a_n}&=&(n-1)^2+a_{n-1}+\dfrac{1}{a_{n-1}},\end{array}$$ and summing the two columns, we get $$\dfrac{1}{a_n}=\dfrac{(n-1)n(2n-1)}{6}+\sum_{k=0}^{n-1} a_k+1.$$
Hence, $$\dfrac{1}{n^3a_n}=\dfrac{(n-1)n(2n-1)}{6n^3}+\dfrac{\sum_{k=0}^{n-1} a_k+1}{n^3}.$$ Since $$\displaystyle \lim_{n \to \infty} \dfrac{\sum_{k=0}^{n-1} a_k+1}{n^3}=\lim_{n \to \infty} \dfrac{c+1}{n^3}=0,$$ we get $$\lim_{n \to \infty} \dfrac{1}{n^3 a_n}=\lim_{n \to \infty} \dfrac{(n-1)n(2n-1)}{6n^3}=\dfrac{1}{3},$$ so $$\lim_{n \to \infty} n^3 a_n=3.$$

Mathematical Reflections 2015, Issue 1 - Problem U326

Problem:
Find $$\sum_{n=0}^\infty \dfrac{a_n+2}{a^2_n+a_n+1},$$ where $a_0>1$ and $a_{n+1}=\dfrac{1}{3}(a^3_n+2)$ for all integers $n \geq 0$.

Proposed by Arkady Alt.

Solution:

Since $a_0>1$, then it can be proved by induction that $a_n>1$ for all $n \in \mathbb{N}$.
For all $n \geq 0$ we have $$\begin{array}{lll} \dfrac{a_n+2}{a^2_n+a_n+1}&=&\dfrac{(a_n+2)(a_n-1)}{a^3_n-1}\\ &=& \dfrac{(a_n+2)(a_n-1)}{3(a_{n+1}-1)}\\ &=&\dfrac{a^2_n+a_n+1-3}{3(a_{n+1}-1)}\\ &=&\dfrac{\frac{a^3_n-1}{a_n-1}-3}{3(a_{n+1}-1)} \\  &=& \dfrac{\frac{3(a_{n+1}-1)}{a_n-1}-3}{3(a_{n+1}-1)} \\ &=&\dfrac{a_{n+1}-a_n}{(a_n-1)(a_{n+1}-1)}\\&=&\dfrac{1}{a_n-1}-\dfrac{1}{a_{n+1}-1}  \end{array}$$
Therefore, $$\sum_{n=0}^\infty \dfrac{a_n+2}{a^2_n+a_n+1}=\dfrac{1}{a_0-1}.$$

Mathematical Reflections 2015, Issue 1 - Problem J325

Problem:
For positive real numbers $a$ and $b$, define their \emph{perfect mean} to be half of the sum of their
arithmetic and geometric means. Find how many unordered pairs of integers $(a,b)$ from the set $\{1,2,\ldots,2015\}$ have their perfect mean a perfect square.

Proposed by Ivan Borsenco.

Solution:

We have $$\dfrac{\frac{a+b}{2}+\sqrt{ab}}{2}=n^2 \qquad n \in \mathbb{N},$$ i.e. $$\sqrt{a}+\sqrt{b}=2n, \qquad n \in \mathbb{N}.$$ As $a,b \in \{1,2,\ldots,2015\}$, then $a$ and $b$ must be perfect squares and $n \leq 44$. Assume that $a \leq b$. If $a$ is an odd perfect square, then $b$ is an odd perfect square, and making a case by case analysis for $a \in \{1,3^2,\ldots,43^2\}$ we get $22+21+\ldots+2+1=253$ unordered pairs. Similarly, if $a$ is an even perfect square, then $b$ is an even perfect square and we get $22+21+\ldots+2+1=253$ unordered pairs. So, there are $253+253=506$ unordered pairs which satisfy the given conditions.