Mathematical Reflections 2017, Issue 1 - Problem O397

Problem:
Solve in integers the equation: $$(x^3-1)(y^3-1)=3(x^2y^2+2).$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
The given equation can be written as $$x^3y^3-(x^3+y^3)-3x^2y^2=5.$$ Let $s=x+y$ and $t=xy$. Observe that $x^3+y^3=(x+y)^3-3xy(x+y)=s^3-3st$, so the given equation becomes $$(t^3-s^3)-3t(t-s)=5,$$ i.e. $$(t-s)(t^2+ts+s^2-3t)=5.$$
We obtain the four systems of equations: $$\begin{array}{rcl} t-s&=&\pm 1 \\ t^2+ts+s^2-3t&=& \pm 5, \end{array} \qquad \begin{array}{rcl} t-s&=&\pm 5 \\ t^2+ts+s^2-3t&=& \pm 1 \end{array}$$
If $t-s=\pm 1$, then $t^2-2ts+s^2=1$ and subtracting this equation to the second equation, we get $3ts-3t=4$ or $3ts-3t=-6$. The first equation is impossible, the second gives $t(s-1)=-2$. So, $(s,t) \in \{(-1,1),(0,2),(2,-2),(3,-1)\}$. It's easy to see that none of these pairs satisfies $t-s=\pm 1$, so there are no solutions in this case. If $t-s=\pm 5$, then $t^2-2ts+s^2=25$ and subtracting this equation to the second equation, we get $3ts-3t=-24$ or $3ts-3t=-26$. The second equation is impossible, the first gives $t(s-1)=-8$. So, $(s,t) \in \{(-7,1),(-3,2),(-1,4),(0,8),(2,-8),(3,-4),(5,-2),(9,-1)\}$. As $t-s=\pm 5$, we obtain $(s,t) \in \{(-3,2),(-1,4)\}$. Since $s$ and $t$ also satisfy the condition $s^2-4t=n^2$ for some $n \in \mathbb{Z}$, we obtain $(s,t)=(-3,2)$. So, $x+y=-3$ and $xy=2$, which gives $(x,y) \in \{(-1,-2),(-2,-1)\}$.

Mathematical Reflections 2017, Issue 1 - Problem U402

Problem:
Let $n$ be a positive integer and let $P(x)$ be a polynomial of degree at most $n$ such that $|P(x)| \leq x+1$ for all $x \in [0,n]$. Prove that
$$|P(n+1)|+|P(-1)| \leq (n+2)(2^{n+1}-1)$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
In order to prove the inequality, we deduce upper bounds for $|P(-1)|$ and $|P(n+1)|$.
Let $a_0,a_1,\ldots,a_n$ be $n+1$ distinct real numbers. Let $$L_i(x)=\prod_{\begin{smallmatrix} j=0\\ j\neq i\end{smallmatrix}}^{n} \dfrac{x-a_j}{a_i-a_j}$$ be the $n$-th degree Lagrange base polynomials for $i=0,1,\ldots,n$. We first prove that $\{L_0(x),\ldots,L_n(x)\}$ is a basis for $\mathbb{R}_n[x]$. Since $\dim \mathbb{R}_n[x]=n+1$, we only have to prove that $\{L_0(x),\ldots,L_n(x)\}$ are linearly independent. Let $(\lambda_0,\ldots,\lambda_{n+1}) \in \mathbb{R}^{n+1}$ such that $$\lambda_0L_0(x)+\ldots+\lambda_n L_n(x)=0$$ For each $i \in \{0,1,\ldots,n\}$, if $x=a_i$ we have $$0=\sum_{j=0}^n \lambda_j L_j(a_i)=\sum_{j=0}^n \lambda_j \delta_{ji}=\lambda_i,$$ so $\lambda_i=0$ for any $i \in \{0,1,\ldots,n\}$ and this shows that $\{L_0(x),\ldots,L_n(x)\}$ is a basis for $\mathbb{R}_n[x]$. So, if $P \in \mathbb{R}_n[x]$, there exists $(\lambda_0,\ldots,\lambda_n) \in \mathbb{R}^{n+1}$ such that $P(x)=\sum_{j=0}^n \lambda_jL_j(x)$. If $x=a_i$, then $$P(a_i)=\sum_{j=0}^n \lambda_j L_j(a_i)=\sum_{j=0}^n \lambda_j \delta_{ji}=\lambda_i,$$ so we can write $$P(x)=\sum_{j=0}^n P(a_j)L_j(x).$$ If $(a_0,a_1,\ldots,a_n)=(0,1,\ldots,n)$, we have $$P(x)=\sum_{j=0}^n P(j)L(x)$$
Now, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |L_i(-1)|=\left| \prod_{\begin{smallmatrix} j=0\\ j\neq i\end{smallmatrix}}^{n} \dfrac{-1-j}{i-j} \right|&=& \displaystyle \prod_{j=0}^{i-1} \dfrac{j+1}{i-j} \prod_{j=i+1}^n \dfrac{j+1}{j-i}\\&=& \displaystyle \dfrac{i!}{i!}\cdot \dfrac{(n+1)!}{(i+1)!(n-i)!}\\&=& \displaystyle {n+1 \choose i+1} \end{array}$$
and
$$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |L_i(n+1)|=\left| \prod_{\begin{smallmatrix} j=0\\ j\neq i\end{smallmatrix}}^{n} \dfrac{n+1-j}{i-j} \right|&=& \displaystyle \prod_{j=0}^{i-1} \dfrac{n+1-j}{i-j} \prod_{j=i+1}^n \dfrac{n+1-j}{j-i}\\&=& \displaystyle \dfrac{(n+1)!}{i!(n+1-i)!}\cdot \dfrac{(n-i)!}{(n-i)!}\\&=& \displaystyle {n+1 \choose i}. \end{array}$$
Hence, $$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |P(-1)|=\left| \sum_{i=0}^n P(i)L_i(-1) \right| & \leq & \displaystyle \sum_{i=0}^n |P(i)L_i(-1)| \\ & \leq & \displaystyle \sum_{i=0}^n (i+1) {n+1 \choose i+1}\\&=& \displaystyle (n+1) \sum_{i=0}^n {n \choose i}\\&=&(n+1)2^n \end{array}$$ and
$$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |P(n+1)|=\left| \sum_{i=0}^n P(i)L_i(n+1) \right| & \leq & \displaystyle \sum_{i=0}^n |P(i)L_i(n+1)| \\ & \leq & \displaystyle \sum_{i=0}^n (i+1) {n+1 \choose i} \\ &=& \displaystyle \sum_{i=0}^n i{n+1 \choose i}+\sum_{i=0}^n {n+1 \choose i} \\ &=& \displaystyle (n+1) \sum_{i=1}^n {n \choose i-1}+ \sum_{i=0}^n {n+1 \choose i} \\&=&(n+1)(2^n-1)+(2^{n+1}-1). \end{array}$$
Adding the last two inequalities, we get the desired inequality.

Mathematical Reflections 2017, Issue 1 - Problem U401

Problem:
Let $P$ be a polynomial of degree $n$ such that $P(k)=\dfrac{1}{k^2}$ for all $k=1,2,\ldots,n+1$. Determine $P(n+2)$.

Proposed by Dorin Andrica, Babe\c{s}-Bolyai University, Cluj-Napoca, Romania

Solution:
There exists a unique interpolating polynomial $P$ of degree $n$ such that $P(k)=\dfrac{1}{k^2}$ for all $k=1,2,\ldots,n+1$ and this is
$$P(x)=\sum_{k=1}^{n+1} \left(\prod_{\stackrel{1\leq j\leq n+1}{j\neq k}}\frac{x-j}{k-j}\right)\dfrac{1}{k^2}.$$
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \prod_{\stackrel{1\leq j\leq n+1}{j\neq k}}\frac{n+2-j}{k-j}&=&\displaystyle\prod_{j=1}^{k-1} \left(\dfrac{n+2-j}{k-j}\right)\prod_{j=k+1}^{n+1} \left(\dfrac{n+2-j}{k-j}\right)\\&=& \displaystyle \dfrac{(n+1)!}{(n-k+2)!(k-1)!}\cdot\dfrac{(n-k+1)!}{(-1)^{n-k+1}(n-k+1)!}\\&=& \displaystyle (-1)^{n-k+1}{n+1 \choose k-1}. \end{array}$$
So, $$P(n+2)=\sum_{k=1}^{n+1} (-1)^{n-k+1}{n+1 \choose k-1}\dfrac{1}{k^2}.$$

Mathematical Reflections 2017, Issue 1 - Problem U397

Problem:
Let $T_n$ be the $n$-th triangular number. Evaluate $$\sum_{n \geq 1} \dfrac{1}{(8T_n-3)(8T_{n+1}-3)}$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $t_n=\dfrac{1}{(8T_n-3)(8T_{n+1}-3)}$.
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{1}{t_n}=(8T_n-3)(8T_{n+1}-3)&=&\left(8\dfrac{n(n+1)}{2}-3\right)\left(8\dfrac{(n+1)(n+2)}{2}-3\right)\\&=&(4n^2+4n-3)(4n^2+12n+5)\\&=&(2n-1)(2n+3)(2n+1)(2n+5). \end{array}$$
We get $$\renewcommand{\arraystretch}{2} \begin{array}{lll} t_n&=&\dfrac{1}{(2n-1)(2n+3)(2n+1)(2n+5)}\\&=&\dfrac{1}{8}\dfrac{1}{(2n-1)(2n+5)}-\dfrac{1}{8}\dfrac{1}{(2n+1)(2n+3)}\\&=&\dfrac{1}{48}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+5}\right)-\dfrac{1}{16}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right)\\&=&\dfrac{1}{48}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+\dfrac{1}{48}\left(\dfrac{1}{2n+3}-\dfrac{1}{2n+5}\right)-\dfrac{1}{24}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right). \end{array}$$
So, $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \sum_{n \geq 1} t_n&=& \displaystyle \dfrac{1}{48}\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+\dfrac{1}{48}\sum_{n=1}^\infty \left(\dfrac{1}{2n+3}-\dfrac{1}{2n+5}\right)-\dfrac{1}{24}\sum_{n=1}^\infty \left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right) \\ &=&\dfrac{1}{48}+\dfrac{1}{48}\cdot\dfrac{1}{5}-\dfrac{1}{24}\cdot\dfrac{1}{3} \\ &=& \dfrac{1}{90}. \end{array}$$

Mathematical Reflections 2017, Issue 1 - Problem S402

Problem:
Prove that $$\sum_{k=1}^{31} \dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}<\dfrac{3}{2}+\sum_{k=1}^{31} (k-1)^{1/5}.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}<\dfrac{k}{3(k-1)^{4/5}},$$ so
$$\begin{array}{lll} \displaystyle \sum_{k=1}^{31} \left(\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}-(k-1)^{1/5}\right)&<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\sum_{k=2}^{31} \dfrac{2k-3}{(k-1)^{4/5}} \\ &<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\sum_{k=2}^{31} \dfrac{2k-3}{k-1}\\&<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\left(1+\dfrac{3}{2}\right)<0. \end{array}$$
So, $$\sum_{k=1}^{31} \left(\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}-(k-1)^{1/5}\right)<0<\dfrac{3}{2}.$$

Mathematical Reflections 2017, Issue 1 - Problem S400

Problem:
Find all $n$ for which $(n-4)!+\dfrac{1}{36n}(n+3)!$ is a perfect square.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly, $n \geq 4$. We have $$\begin{array}{lll} (n-4)!+\dfrac{1}{36n}(n+3)!&=&(n-4)!\left(1+\dfrac{1}{36n}(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)\right)\\&=&(n-4)!\left(1+\dfrac{1}{36}(n^2-9)(n^2-4)(n^2-1)\right)\\&=&(n-4)!\left(\dfrac{n^6-14n^4+49n^2}{36}\right)\\&=&(n-4)!\left(\dfrac{n(n^2-7)}{6}\right)^2. \end{array}$$
Observe that $n(n^2-7)=n^3-7n \equiv n^3-n \equiv 0 \pmod{6}$, so $\dfrac{n(n^2-7)}{6}$ is an integer. It follows that $(n-4)!\left(\dfrac{n(n^2-7)}{6}\right)^2$ is a perfect square if and only if $(n-4)!$ is a perfect square. If $n=4$ or $n=5$, then $(n-4)!=1$, which is a perfect square. Let $n>5$ and let $p$ be the greatest prime that divides $(n-4)!$. By Bertrand's Postulate, there exists a prime $q$ such that $p<q<2p$. If $2p \leq n-4$, then $q<n-4$, which gives $q \ | \ (n-4)!$, contradiction. So, $n-4<2p$, which means that $p \ | \ (n-4)!$ and $p^2 \nmid (n-4)!$. So, $(n-4)!$ is not a perfect square if $n>5$. Therefore, $n \in \{4,5\}$.

Mathematical Reflections 2017, Issue 1 - Problem S397

Problem:
Let $a,b,c$ be positive real numbers. Prove that $$\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{3(ab+bc+ca)}{2(a+b+c)} \geq a+b+c.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
The given inequality is equivalent to $$(a+b+c)\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)+\dfrac{3}{2}(ab+bc+ca) \geq (a+b+c)^2,$$ i.e.

$$\dfrac{a^2c}{a+b}+\dfrac{b^2a}{b+c}+\dfrac{c^2b}{c+a} \geq \dfrac{1}{2}(ab+bc+ca) \qquad (1)$$

By the AM-GM Inequality, we have $$\dfrac{2a^2c}{a+b}+\dfrac{c(a+b)}{2} \geq 2ca,$$ $$\dfrac{2b^2a}{b+c}+\dfrac{a(b+c)}{2} \geq 2ab,$$ $$\dfrac{2c^2b}{c+a}+\dfrac{b(c+a)}{2} \geq 2bc.$$ Adding these three inequalities, we get inequality (1). The equality holds if and only if $a=b=c$.

Mathematical Reflections 2017, Issue 1 - Problem J401

Problem:
Find all integers $n$ for which $n^2+2^n$ is a perfect square.

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
Clearly, $n \geq 0$. If $n=0$, we get $n^2+2^n=1$, which is a perfect square. Let $n>0$. If $n$ is even, then $n=2k$ for some $k \in \mathbb{N}^*$. If $k \geq 7$, then $$(2^k)^2=2^{2k}<4k^2+2^{2k}<2^{2k}+2^{k+1}+1=(2^k+1)^2,$$ so $n^2+2^n$ is not a perfect square if $n$ is even and $n \geq 14$. So, $n \in \{2,4,6,8,10,12\}$. An easy check gives the solution $n=6$. If $n$ is odd, then $n=2k+1$ for some $k \in \mathbb{N}$. If $k=0$, we get no solutions, so assume $k \geq 1$. Let $m \in \mathbb{N}^*$ such that $(2k+1)^2+2^{2k+1}=m^2$. Then, $$(m-2k-1)(m+2k+1)=2^{2k+1}.$$ Since $m-2k-1<m+2k+1$ and the two factors have the same parity, then $$\begin{array}{lll} m-2k-1&=&2^a \\ m+2k+1&=&2^b, \end{array}$$ where $a,b \in \mathbb{N}$, $1 \leq a \leq b \leq 2k$ and $a+b=2k+1$. If $a \geq 2$, then subtracting we get $2(2k+1)=2^b-2^a=2^a(2^{b-a}-1)$, i.e. $2k+1=2^{a-1}(2^{b-a}-1)$, contradiction. So, $a=1$ and $b=2k$, which gives $2k+1=2^{2k-1}-1$, i.e. $k=2^{2k-2}-1$. If $k \geq 2$, then $k<2^{2k-2}-1$, so it must be $k=1$. But if $k=1$, we get no solutions. So, there are no solutions when $n$ is odd. We conclude that $n \in \{0,6\}$.

Mathematical Reflections 2017, Issue 1 - Problem J400

Problem:
Prove that for all real numbers $a,b,c$ the following inequality holds:
$$\dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|} \geq \dfrac{|a+b+c|}{1+|a+b+c|}.$$
When does the equality occur?

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
 Put $s=a+b+c$. By Triangle Inequality, we have $|s| \leq |a|+|b|+|c|$, so $|s|(1+|a|+|b|+|c|) \leq (1+|s|)(|a|+|b|+|c|)$, i.e.
$$\dfrac{|s|}{1+|s|} \leq \dfrac{|a|+|b|+|c|}{1+|a|+|b|+|c|}.$$ Using the fact that $|x| \geq 0$ for all real numbers $x$, we have
$$\begin{array}{lll} \dfrac{|a|+|b|+|c|}{1+|a|+|b|+|c|}&=&\dfrac{|a|}{1+|a|+|b|+|c|}+\dfrac{|b|}{1+|a|+|b|+|c|}+\dfrac{|c|}{1+|a|+|b|+|c|} \\ & \leq & \dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|}. \end{array}$$
So, $$\dfrac{|s|}{1+|s|} \leq \dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|},$$ which is the desired inequality. Equality occurs if and only if $|a|=|b|=|c|=0$, i.e. if and only if $a=b=c=0$.

Mathematical Reflections 2017, Issue 1 - Problem J399

Problem:
Two nine-digit numbers $m$ and $n$ are called cool if

   (a) they have the same digits but in different order,
   (b) no digit appears more than once,
   (c) $m$ divides $n$ or $n$ divides $m$.

Prove that if $m$ and $n$ are cool, then they contain digit $8$.

Proposed by Titu Andreescu, Dallas, Texas

Solution:
Assume by contradiction that there exist two \emph{cool} numbers $m$ and $n$ not containing digit $8$. Then in $m$ and $n$ appear the digits $0,1,2,3,4,5,6,7,9$ exactly once. Since the sum of their digits is $37$, then $m,n \equiv 1 \pmod{9}$. Assume without loss of generality that $m$ divides $n$. Then, $n=mk$, where $k$ is a natural number. Hence, $n-m=m(k-1)$. Reducing modulo $9$ this equation, we get $k-1 \equiv 0 \pmod{9}$, i.e. $k-1$ is divisible by $9$. Since $m$ and $n$ have the same digits in different order, then $m \neq n$, which gives $k \neq 1$. So, $k \geq 10$. But then $n \geq 10m$, i.e. $n$ has more digits than $m$, contradiction.

Mathematical Reflections 2017, Issue 1 - Problem J398

Problem:
Let $a,b,c$ be real numbers. Prove that $$(a^2+b^2+c^2-2)(a+b+c)^2+(1+ab+bc+ca)^2 \geq 0.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Let $x=a+b+c$ and $y=ab+bc+ca$. Then, $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y.$$ So,
$$\begin{array}{lll} (a^2+b^2+c^2-2)(a+b+c)^2+(1+ab+bc+ca)^2&=&(x^2-2y-2)x^2+(1+y)^2\\&=&x^4-2x^2(y+1)+(y+1)^2\\&=&(x^2-y-1)^2 \\ & \geq & 0. \end{array}$$
Equality holds if and only if $x^2=y+1$, i.e. if and only if $(a+b+c)^2=ab+bc+ca+1$, i.e. if and only if $a^2+b^2+c^2=1-ab-bc-ca$.

Mathematical Reflections 2017, Issue 1 - Problem J397

Problem:
Find all positive integers $n$ for which $3^4+3^5+3^6+3^7+3^n$ is a perfect square.

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
We have $3^4+3^5+3^6+3^7+3^n=3240+3^n$. If $n \leq 4$, we obtain that $3240+3^n$ is a perfect square only when $n=2$. Assume that $n>4$. Then, there exists $t \in \mathbb{N}$ such that $3240+3^n=t^2$, i.e. $$3^4(40+3^{n-4})=t^2.$$ Since $3^4$ is a perfect square, it follows that $40+3^{n-4}$ must be a perfect square, so there exists $m \in \mathbb{N}$ such that $40+3^{n-4}=m^2$. If $n-4$ is odd, then $40+3^{n-4} \equiv 3 \pmod{4}$, so it cannot be a perfect square. It follows that $n-4$ is even, i.e. $n-4=2k$, where $k \in \mathbb{N}$, $k>2$. So, $40+3^{2k}=m^2$, i.e. $$(m-3^k)(m+3^k)=40.$$ Since $m-3^k<m+3^k$ and both factors have the same parity, then we obtain the two systems of equations
$$\begin{array}{lll} m-3^k&=&2 \\ m+3^k&=&20, \end{array} \qquad \begin{array}{lll} m-3^k&=&4 \\ m+3^k&=&10. \end{array}$$
Subtracting the first equation to the second equation of each system, we get $3^k=9$ or $3^k=3$, i.e. $k=2$ or $k=1$. So, $n=8$ or $n=6$. We conclude that $n \in \{2,6,8\}$.

Mathematical Reflections 2016, Issue 6 - Problem O391

Problem:
Find all 4-tuples $(x,y,z,w)$ of positive integers such that $$(xy)^3+(yz)^3+(zw)^3-252yz=2016.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
The given equation can be written as $$(xy)^3+(zw)^3+yz(y^2z^2-252)=2016.$$ Observe that it must be $yz(y^2z^2-252) \leq 2016$, so $yz \leq 18$. It follows that $$\begin{array}{rcl} (xy)^3+(zw)^3 & \in & \{2267,2512,2745,2960,3151,3312,3437,3520,3555,\\ & & 3536,3457,3312,3095,2800,2421,1952,1387,720\}. \end{array}$$
A case by case analysis gives $(xy)^3+(zw)^3 \in \{2745,2960\}$. If $(xy)^3+(zw)^3=2745$, we get $xy=1,zw=14,yz=3$ or $xy=14,zw=1,yz=3$, which gives no solutions. If $(xy)^3+(zw)^3=2960$, then $xy=6,zw=14,yz=4$ or $xy=14,zw=6,yz=4$. We get $(x,y,z,w) \in \{(3,2,2,7),(7,2,2,3)\}$.

Mathematical Reflections 2016, Issue 6 - Problem U395

Problem:
Evaluate $\displaystyle \int \dfrac{x^2+6}{(x \cos x-3\sin x)^2} \ dx$.

Proposed by Abdelouahed Hamdi, Doha, Qatar

Solution:Observe that $$\begin{array}{lll} \dfrac{x^2+6}{(x \cos x-3\sin x)^2}&=&\dfrac{x^2(\cos^2 x +\sin^2 x)+6(\cos^2 x + \sin^2 x)+5 x \sin x \cos x - 5x \sin x \cos x}{(x \cos x-3\sin x)^2} \\ &=& \dfrac{(x \cos x-2\sin x)(x \cos x-3 \sin x)+(x \sin x+3 \cos x)(x \sin x+2\cos x)}{(x\cos x-3\sin x)^2}\\&=& \dfrac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2} \\ &=&\left(\dfrac{f(x)}{g(x)}\right)', \end{array}$$ where $f(x)=x\sin x+3\cos x$ and $g(x)=x \cos x -3\sin x$. So, $$\int \dfrac{x^2+6}{(x \cos x-3\sin x)^2} \ dx=\dfrac{x\sin x+3\cos x}{x \cos x-3\sin x}+C.$$

Mathematical Reflections 2016, Issue 6 - Problem U391

Problem:
Find all positive integers $n$ such that $\varphi^3(n) \leq n^2$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:

Clearly $n=1$ satisfies the condition. Let $n>1$ and let $n=\prod_{j=1}^m p_j^{k_j}$, where $p_j$ is the $j$-th prime and $k_j \in \mathbb{N}$ for $j=1,2,\ldots,m$. Since $f(n)=n^2$ and $\varphi(n)$ are multiplicative functions, then the given relation becomes $$\prod_{j=1}^m \varphi^3(p_j^{k_j}) \leq \prod_{j=1}^m p_j^{2k_j}.$$ We use the following Lemma.


Lemma.

Let $p$ be a prime number and let $k$ be a positive integer.

    (a) If $k=0$, then $\varphi^3(p^k)=p^{2k}$.
    (b) If $p \geq 5$, then
    $$\varphi^3(p^k)>\dfrac{9}{4}p^{2k} \qquad \forall k \in \mathbb{N}^*. \qquad (1)$$
    (c) If $p \geq 7$, then $$\varphi^3(p^k)>4p^{2k} \qquad \forall k \in \mathbb{N}^*. \qquad (2)$$
    (d) If $p \geq 11$, then  $$\varphi^3(p^k)>8p^{2k} \qquad \forall k \in \mathbb{N}^*.  \qquad (3)$$

Proof. It's a simple computation.


From the inequality (1) in the Lemma we must consider only the primes $p_1=2$ and $p_2=3$.
If $k_1>3$, then $$\varphi^3(2^{k_1})=(2^{k_1}-2^{k_1-1})^3=2^{3(k_1-1)} > 2^{2k_1}$$ and if $k_2>1$, then $$\varphi^3(3^{k_2})=(3^{k_2}-3^{k_2-1})^3=3^{3(k_2-1)}\cdot8 > 3^{2k_2}.$$ So, there are no solutions if $k_1=k_2=0$ and $k_j>0$ for some $j>2$ or $k_1>3$ and $k_2>1$. So, $k_1 \leq 3$ or $k_2 \leq 1$.
\begin{description}

(a) If $k_1=0$, then $n=\prod_{j=2}^m p_j^{k_j}$. As before, we conclude immediately that there are solutions only if $k_2 \leq 1$ and $k_j=0$ for all $j=1,2,\ldots,m$. So, $n=3$.

(b) If $k_1=1$, then $n=2a$, where $a$ is an odd number. Hence, $$\varphi^3(n)=\varphi^3(2a)=\varphi^3(2)\varphi^3(a)=\varphi^3(a).$$ We have to find $a$ odd such that $$\varphi^3(a) \leq 4a^2.$$ From the inequality (2) in the Lemma we must consider only the primes $p_2=3$ and $p_3=5$. If $k_2>2$, then $$\varphi^3(3^{k_2})=3^{3(k_2-1)}\cdot2^3>4\cdot3^{2k_2}$$ and if $k_3>1$, then $$\varphi^3(5^{k_3})=5^{3(k_3-1)}\cdot4^3>4\cdot5^{2k_3}$$ So, there are no solutions for $a$ if $k_2=k_3=0$ and $k_j>0$ for some $j>3$ or $k_2>2$ and $k_3>1$. It follows that $k_2 \leq 2$ or $k_3 \leq 1$.

  (i) If $k_2=0$, then we have a solution if $k_3 \leq 1$, so $a \in \{1,5\}$, which gives $n \in \{2,10\}$.

  (ii) If $k_2=1$, then $a=3b$, where $b$ is a natural number nondivisible by $2$ and $3$. Hence, $n=6b$ and $$\varphi^3(n)=\varphi^3(6b)=\varphi^3(6)\varphi^3(b)=8\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $3$ such that $8\varphi^3(b) \leq 36b^2$, i.e. $$\varphi^3(b) \leq \dfrac{9}{2}b^2.$$ From the inequality (3) in the Lemma we must consider only the primes $p_3=5$ and $p_4=7$. If $k_3>1$, then $$\varphi^3(5^{k_3})=5^{3(k_3-1)}\cdot4^3>\dfrac{9}{2}\cdot5^{2k_3}$$ and if $k_4>1$, then $$\varphi^3(7^{k_4})=7^{3(k_4-1)}\cdot6^3>\dfrac{9}{2}\cdot7^{2k_4}$$ So, there are no solutions for $b$ if $k_3=k_4=0$ and $k_j>0$ for some $j>4$ or $k_3>1$ and $k_4>1$. It follows that $k_3 \leq 1$ or $k_4 \leq 1$. If $k_3=0$, then we have a solution if $k_4 \leq 1$, so $b \in \{1,7\}$, which gives $n \in \{6,42\}$. If $k_3=1$, then $b=5c$, where $c$ is a natural number nondivisible by $2,3,5$. Hence, $n=30c$ and $$\varphi^3(n)=\varphi^3(30c)=\varphi^3(30)\varphi^3(c)=512\varphi^3(c).$$ We have to find $c$ nondivisible by $2,3,5$ such that $512\varphi^3(c) \leq 900c^2$, i.e. $$\varphi^3(c) \leq \dfrac{225}{128}c^2.$$ From the inequality (2) in the Lemma we have no primes for $c$. So, $c=1$ and $n=30$.

  (iii) If $k_2=2$, then $a=9b$, where $b$ is a natural number nondivisible by $2$ and $3$. Hence, $n=18b$ and $$\varphi^3(n)=\varphi^3(18b)=\varphi^3(18)\varphi^3(b)=216\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $3$ such that $216\varphi^3(b) \leq 324b^2$, i.e. $$\varphi^3(b) \leq \dfrac{3}{2}b^2.$$ From the inequality (1) in the Lemma we have no primes for $b$. So, $b=1$ and $n=18$.

  (iv) If $k_3=0$, then we have a solution if $k_2 \leq 2$, so $a \in \{1,3,9\}$, which gives $n \in \{2,6,18\}$.

  (v) If $k_3=1$, then $a=5b$, where $b$ is a natural number nondivisible by $2$ and $5$. Hence, $n=10b$ and $$\varphi^3(n)=\varphi^3(10b)=\varphi^3(10)\varphi^3(b)=64\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $5$ such that $64\varphi^3(b) \leq 100b^2$, i.e. $$\varphi^3(b) \leq \dfrac{25}{16}b^2.$$ From the inequality (2) in the Lemma we must consider only the prime $p_2=3$. If $k_2>1$, then $$\varphi^3(3^{k_2})=3^{3(k_2-1)}\cdot2^3>\dfrac{25}{16}\cdot3^{2k_2}.$$ So, there are no solutions for $b$ if $k_2=0$ and $k_j>0$ for some $j>3$ or $k_2>1$. It follows that $k_2 \leq 1$. If $k_2=0$, then $b=1$, which gives $n=10$. If $k_2=1$, then $b=3c$, where $c$ is a natural number nondivisible by $2,3,5$. Hence, $n=30c$ and we conclude as in point (ii).

(c) If $k_1=2$, then $n=4a$, where $a$ is an odd number. Hence, $$\varphi^3(n)=\varphi^3(4a)=\varphi^3(4)\varphi^3(a)=8\varphi^3(a).$$ We have to find $a$ odd such that $8\varphi^3(a) \leq 16a^2$, i.e. $$\varphi^3(a) \leq 2a^2.$$ From the inequality (1) in the Lemma we must consider only the prime $p_2=3$.  If $k_2>1$, then $$\varphi^3(3^{k_2})=3^{3(k_2-1)}\cdot2^3>2\cdot3^{2k_3}.$$ So, there are no solutions for $a$ if $k_2=0$ and $k_j>0$ for some $j>2$ or $k_2>1$. It follows that $k_2 \leq 1$. If $k_2=0$, then $a=1$ and $n=4$. If $k_2=1$, then $a=3b$, where $b$ is a natural number nondivisible by $2$ and $3$. Hence, $n=12b$ and $$\varphi^3(n)=\varphi^3(12b)=\varphi^3(12)\varphi^3(b)=64\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $3$ such that $64\varphi^3(b) \leq 144b^2$, i.e. $$\varphi^3(b) \leq \dfrac{9}{4}b^2.$$ From the inequality (1) in the Lemma we have no primes for $b$. So, $b=1$ and $n=12$.

(d) If $k_1=3$, then $n=8a$, where $a$ is an odd number. Hence, $$\varphi^3(n)=\varphi^3(8a)=\varphi^3(8)\varphi^3(a)=64\varphi^3(a).$$ We have to find $a$ odd such that $64\varphi^3(a) \leq 64a^2$, i.e. $$\varphi^3(a) \leq a^2.$$ We know that this inequality has a solution if $k_2 \leq 1$, so $a \in \{1,3\}$, which gives $n \in \{8,24\}$.

(e) If $k_2=0$, then $n=\prod_{\substack{j=1 \\ j \neq 2}}^m p_j^{k_j}$. As before, we conclude immediately that there are solutions only if $k_1 \leq 3$ and $k_j=0$ for all $j=1,2,\ldots,m$. So, $n \in \{2,4,8\}$.

(f) If $k_2=1$, then $n=3a$, where $a$ is a natural number nondivisible by $3$. Hence, $$\varphi^3(n)=\varphi^3(3a)=\varphi^3(3)\varphi^3(a)=8\varphi^3(a).$$ We have to find $a$ nondivisible by $3$ such that $8\varphi^3(a) \leq 9a^2$, i.e. $$\varphi^3(a) \leq \dfrac{9}{8}a^2.$$ From the inequality (1) in the Lemma we must consider only the prime $p_1=2$. If $k_1>2$, then $$\varphi^3(2^{k_1})=2^{3(k_1-1)}>\dfrac{9}{8}\cdot2^{2k_1}.$$ So, there are no solutions for $a$ if $k_1$ and $k_j>0$ for some $j>2$ or $k_1>2$. It follows that $k_1 \leq 2$, so $a \in \{1,2,4\}$, which gives $n \in \{1,6,12\}$.

In conclusion, we have $n \in \{1,2,3,4,6,8,10,12,18,24,30,42\}$.

Note: Let $$A_x=\{n \in \mathbb{N}^* \ | \ \varphi(n) \leq n^x\}.$$ Observe that the function $$g(x)=|A_x|$$ is increasing for $x \in [0,1)$. The problem tells us that $g(2/3)=12$. Moreover, it's easy to see that $g(0)=2$ and $g(1/2)=4$.

Mathematical Reflections 2016, Issue 6 - Problem S395

Problem:
Let $a,b,c$ be positive integers such that $$a^2b^2+b^2c^2+c^2a^2-69abc=2016.$$
Find the least possible value of $\min(a,b,c)$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume without loss of generality that $a \leq b \leq c$. If $a=1$, then $$b^2+b^2c^2+c^2-69bc=2016.$$ This equation is equivalent to
$$(2bc-67)^2+4(c-b)^2=12553.$$ It follows that $0 \leq c-b \leq 56$ and $0 < bc \leq 89$. Hence $b^2 \leq 89$, which gives $b \leq 9$. An easy check gives no positive integer solutions for $c$. If $a=2$, then $$4b^2+b^2c^2+4c^2-138bc=2016.$$
This equation is equivalent to $$(bc-65)^2+4(c-b)^2=6241.$$ It follows that $0 \leq c-b \leq 39$ and $0 < bc \leq 144$. Hence $b^2 \leq 144$, which gives $b \leq 12$. An easy check gives $b=c=12$, so the least possible value of $\min(a,b,c)$ is $2$.

Mathematical Reflections 2016, Issue 6 - Problem S393

Problem:
If $n$ is an integer such that $n^2+11$ is a prime, prove that $n+4$ is not a perfect cube.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume that $n+4$ is a perfect cube. Then $n+4=m^3$, where $m \in \mathbb{Z}$, i.e. $n=m^3-4$. It follows that
$$\begin{array}{lll} n^2+11&=&(m^3-4)^2+11\\&=&m^6-8m^3+27\\&=&m^6+m^3+27-9m^3\\&=&(m^2)^3+(m)^3+(3)^3-9m^3\\&=&(m^2+m+3)(m^4-m^3-2m^2-3m+9). \end{array}$$ As $m^2+m+3 \geq 3$ and $m^4-m^3-2m^2-3m+9 \geq 3$ for all $m \in \mathbb{Z}$, then  $n^2+11$ is not prime, contradiction.

Mathematical Reflections 2016, Issue 6 - Problem J331

Problem:
Solve the equation $$4x^3+\dfrac{127}{x}=2016.$$

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
The given equation is equivalent to $$4x^4-2016x+127=0.$$ Observe that
$$\begin{array}{lll} 4x^4-2016x+127&=&4x^4-256x^2+(2x^2+16x)+127(2x^2-16x)+127\\&=&(2x^2-16x+1)(2x^2+16x)+127(2x^2-16x+1) \\ &=&(2x^2-16x+1)(2x^2+16x+127)\end{array}$$
So, the given equation becomes $$(2x^2-16x+1)(2x^2+16x+127)=0.$$
We obtain $$x=\dfrac{8 \pm \sqrt{62}}{2}, \qquad x=\dfrac{-8 \pm i\sqrt{190}}{2}.$$

Recreatii Matematice 2/2016, Problem VII.208

Problem:
Prove that the number $$N=2016^{n+1}-2015n-2016$$ has at least $27$ divisors for any $n \in \mathbb{N}^*$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
We have $$\begin{array}{lll} N&=&2016\cdot(2016^n-1)-2015n\\&=&2016\cdot2015\cdot(2016^{n-1}+2016^{n-2}+\ldots+1)-2015n\\&=&2015\cdot(2016^n+2016^{n-1}+\ldots+2016-n)\\&=&2015\cdot[(2016^n-1)+(2016^{n-1}-1)+\ldots+(2016-1)] \end{array}$$
Since $2016^n-1$ is divisible by $2015$ for all $n \geq 1$, we have that $2015^2 \ | \ N$. As $2015=5\cdot13\cdot31$, then $2015=5^2\cdot13^2\cdot31^2$, which has $27$ divisors. The conclusion follows.

Gazeta Matematica 6-7-8/2016, Problem 27253

Problem:
Evaluate $$\sum_{n=6}^{\infty} \dfrac{n^3-12n^2+47n-60}{n^5-5n^3+4n}$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$n^3-12n^2+47n-60=(n-5)(n-4)(n-3)$$ and $$n^5-5n^3+4n=(n-2)(n-1)n(n+1)(n+2),$$ so we have to evaluate
$$\sum_{n=6}^{\infty} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}.$$
Set $f(x)=\dfrac{(x-5)(x-4)(x-3)}{(x-2)(x-1)x(x+1)(x+2)}$. Since $x=0,\pm 1,\pm 2$ are simple poles, then $$\textrm{Res}(f,2)=\lim_{x \to 2}(x-2)f(x)=-\dfrac{1}{4},$$ $$\textrm{Res}(f,1)=\lim_{x \to 1}(x-1)f(x)=4,$$ $$\textrm{Res}(f,0)=\lim_{x \to 0}xf(x)=-15,$$ $$\textrm{Res}(f,-1)=\lim_{x \to -1}(x+1)f(x)=20,$$ $$\textrm{Res}(f,-2)=\lim_{x \to -2}(x+2)f(x)=-\dfrac{35}{4}.$$
So, $$\begin{array}{lll} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}&=&-\dfrac{1}{4(n-2)}+\dfrac{4}{n-1}-\dfrac{15}{n}+\dfrac{20}{n+1}-\dfrac{35}{4(n+2)}\\  &=&\left(-\dfrac{1}{4(n-2)}+\dfrac{1}{4(n+2)}\right)+\left(\dfrac{4}{n-1}-\dfrac{4}{n+1}\right)+\\&+&\left(-\dfrac{15}{n}+\dfrac{15}{n+1}\right)+\left(\dfrac{9}{n+1}-\dfrac{9}{n+2}\right). \end{array}$$
Since $$\sum_{n=6}^{\infty} \left(-\dfrac{1}{4(n-2)}+\dfrac{1}{4(n+2)}\right)=\dfrac{1}{4}\left(-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right),$$
$$\sum_{n=6}^{\infty} \left(\dfrac{4}{n-1}-\dfrac{4}{n+1}\right)=4\left(\dfrac{1}{5}+\dfrac{1}{6}\right),$$ $$\sum_{n=6}^{\infty} \left(-\dfrac{15}{n}+\dfrac{15}{n+1}\right)=-\dfrac{15}{6},$$ $$\sum_{n=6}^{\infty} \left(\dfrac{9}{n+1}-\dfrac{9}{n+2}\right)=-\dfrac{9}{8},$$ we have $$\sum_{n=6}^{\infty} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}=\dfrac{1}{4}\left(-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)+4\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\dfrac{15}{6}-\dfrac{9}{8}=\dfrac{1}{16}.$$

Gazeta Matematica 6-7-8/2016, Problem 27248

Problem:
Solve in real numbers the equation: $$\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}=x^2-x-1.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$\begin{array}{lll} (x^6-1)-(3x^5-5x^3+3x)&=&x^6-3x^5+5x^3-3x-1\\&=&(x^2-x-1)(x^4-2x^3-x^2+2x+1)\\&=&(x^2-x-1)(x^2-x-1)^2\\&=&(x^2-x-1)^3. \end{array}$$ Then, from the identity $(a-b)^3=a^3-b^3-3ab(a-b)$, the given equation can be written as
$$(x^6-1)-(3x^5-5x^3+3x)-3\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot[\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}]=(x^2-x-1)^3,$$ i.e. $$\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot[\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}]=0.$$ Since $\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}=x^2-x-1$, then the given equation becomes $$\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot(x^2-x-1)=0,$$ i.e. $$(x^6-1)(3x^5-5x^3+3x)(x^2-x-1)=0.$$ Finally, we have $$(x-1)(x^2+x+1)(x+1)(x^2-x+1)x(3x^4-5x^2+3)(x^2-x-1)=0.$$
Solving each equation, we get $x \in \left\{-1,0,1,\dfrac{1-\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2}\right\}$.

Gazeta Matematica 6-7-8/2016, Problem 27246

Problem:
Do there distinct prime numbers $p,q,r$ such that $\sqrt{p},\sqrt{q},\sqrt{r}$ are terms of an arithmetic progression?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
The answer is no. Assume by contradiction that $\sqrt{p},\sqrt{q},\sqrt{r}$ are in the same arithmetic progression. If $d \neq 0$ is the common difference, then there exist integers $m,n \neq 0$ such that $$\begin{array}{rcl}\sqrt{q}-\sqrt{p}&=&md, \\ \sqrt{r}-\sqrt{p}&=&nd. \end{array}$$ Dividing the first and the second equation by $m$ and $n$ respectively and substituting, we obtain $$m(\sqrt{r}-\sqrt{p})=n(\sqrt{q}-\sqrt{p}),$$ whence $$m\sqrt{r}-n\sqrt{q}=(m-n)\sqrt{p}.$$ Squaring both sides, we obtain $$rm^2+qn^2-2mn\sqrt{rq}=p(m-n)^2,$$ which gives $$\sqrt{rq}=\dfrac{rm^2+qn^2-p(m-n)^2}{2mn}.$$ But this equality is false, since the left-hand side is irrational and the right-hand side is rational.

Crux Mathematicorum 2016, Issue 1 - Problem 4108

Problem:
Write $2010$ as a sum of consecutive squares. Is it possible to write $2014$ as the sum of several consecutive squares?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
(a) Let $k$ be the number of consecutive perfect squares which satisfy the conditions and let $n \in \mathbb{N}$.  Then, $$(n+1)^2+(n+2)^2+\ldots+(n+k)^2=2010.$$ Since $$\begin{array}{lll}(n+1)^2+\ldots+(n+k)^2&=&kn^2+2n(1+2+\ldots+k)+(1^2+2^2+\ldots+k^2)\\&=&kn^2+2n\cdot\dfrac{k(k+1)}{2}+\dfrac{k(k+1)(2k+1)}{6}\\&=&\dfrac{k}{6}[6n^2+6n(k+1)+(k+1)(2k+1)], \end{array}$$ we get
$$k[6n^2+6n(k+1)+(k+1)(2k+1)]=12060.    \qquad          (1)$$
Moreover, since $\dfrac{k(k+1)(2k+1)}{6}=1^2+2^2+\ldots+k^2 \leq (n+1)^2+(n+2)^2+\ldots+(n+k)^2=2010$, we get $k \leq 17$. So, $k \ | \ 12060$ and $k \leq 17$, which gives $k \in \{1,2,3,4,5,6,9,10,12,15\}$. Observe that if $k=4h+2$, where $h \in \mathbb{N}$, then the left-hand side in (1) is not divisible by $4$, but the right-hand side is divisible by $4$. So, $k \in \{1,3,5,9,12,15\}$. If $k=15$, equation (1) gives $$6n(n+16)+496=804 \implies 6n(n+16)=308,$$ a contradiction. If $k=12$, equation (1) gives $$6n(n+13)+325=1005 \implies 6n(n+13)=680,$$ a contradiction. If $k=9$, equation (1) gives $$6n(n+10)+190=1340 \implies 6n(n+10)=1150,$$ a contradiction. If $k=5$, equation (1) gives $$6n(n+6)+66=2412 \implies n(n+6)=391,$$ which gives $n=17$. Therefore, $18^2+19^2+20^2+21^2+22^2=2010$, and the maximum number of consecutive perfect squares which satisfy the condition is $5$.

(b) The answer is no. Indeed, if there exist $k,n \in \mathbb{N}$ such that $$(n+1)^2+(n+2)^2+\ldots+(n+k)^2=2014,$$ then as we have seen in (a),
$$k[6n^2+6n(k+1)+(k+1)(2k+1)]=2014\cdot6=12084   \qquad        (2)$$
and $\dfrac{k(k+1)(2k+1)}{6} \leq 2014$. So, $k \ | \ 12084$, $k \leq 17$ and $k \neq 4h+2$ for any $h \in \mathbb{N}$. It follows that $k \in \{1,3,4,12\}$. Since $2014$ is not a perfect square, $k \neq 1$. As $24^2+25^2+26^2=1877<2014<25^2+26^2+27^2=2030$, then $k \neq 3$. As $20^2+21^2+22^2+23^2=1854<2014<21^2+22^2+23^2+24^2=2030$, then $k \neq 4$. Finally, if $k=12$ equation (2) gives $6n(n+13)+325=1007$, i.e. $6n(n+13)=682$, a contradiction. Therefore there does not exist $k \in \mathbb{N}$ which satisfy the condition, and the conclusion follows.