Mathematical Reflections 2016, Issue 1 - Problem O361

Problem:
Determine the smallest natural number $n>2$ such that there exist $n$ consecutive integers whose sum of their squares is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$(-4)^2+(-3)^2+(-2)^2+(-1)^2+0^2+1^2+2^2+3^2+4^2+5^2+6^2=121=11^2,$$ so $n=11$ works. We prove that this is the smallest natural number greater than $2$ that satisfies the given condition.

(i) $n=3$. Let $m \in \mathbb{Z}$. We have $$(m-1)^2+m^2+(m+1)^2=3m^2+2 \equiv 2 \pmod{3},$$ so the sum of the squares of $3$ consecutive integers cannot be a perfect square.

(ii) $n=4$. Let $m \in \mathbb{Z}$. We have $$(m-1)^2+m^2+(m+1)^2+(m+2)^2=4m^2+4m+6 \equiv 2 \pmod{4},$$ so the sum of the squares of $4$ consecutive integers cannot be a perfect square.

(iii) $n=5$. Let $m \in \mathbb{Z}$. We have $$(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2=5(m^2+2),$$ but $m^2+2 \not \equiv 0 \pmod{5}$, so the sum of the squares of $5$ consecutive integers cannot be a perfect square.

(iv) $n=6$. Let $m \in \mathbb{Z}$. We have $$(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2+(m+3)^2=6m(m+1)+19.$$ Since $m(m+1)$ is even, then $6m(m+1) \equiv 0 \pmod{4}$ and $6m(m+1)+19 \equiv 3 \pmod{4}$. So the sum of the squares of $6$ consecutive integers cannot be a perfect square.

(v) $n=7$. Let $m \in \mathbb{Z}$. We have $$(m-3)^2+(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2+(m+3)^2=7(m^2+4),$$ but $m^2+4 \not \equiv 0 \pmod{7}$, so the sum of the squares of $7$ consecutive integers cannot be a perfect square.

(vi) $n=8$. Let $m \in \mathbb{Z}$. We have $$(m-3)^2+(m-2)^2+\ldots+(m+3)^2+(m+4)^2=4[2m(m+1)+11].$$ Since $m(m+1)$ is even, then $2m(m+1) \equiv 0 \pmod{4}$ and $2m(m+1)+11 \equiv 3 \pmod{4}$. So the sum of the squares of $8$ consecutive integers cannot be a perfect square.

(vii) $n=9$. Let $m \in \mathbb{Z}$. We have $$(m-4)^2+(m-3)^2+\ldots+(m+3)^2+(m+4)^2=3(3m^2+20),$$ but $3m^2+20 \not \equiv 0 \pmod{3}$, so the sum of the squares of $9$ consecutive integers cannot be a perfect square.

(viii) $n=10$. Let $m \in \mathbb{Z}$. We have $$(m-4)^2+(m-3)^2+\ldots+(m+4)^2+(m+5)^2=5[2m(m+1)+17],$$ but $2m(m+1)+17 \not \equiv 0 \pmod{5}$, so the sum of the squares of $10$ consecutive integers cannot be a perfect square.

The conclusion follows.

Note:
The solution is not unique. For example, $$18^2+19^2+20^2+21^2+22^2+23^2+24^2+25^2+26^2+27^2+28^2=77^2.$$
It's easy to see why. We must find $m \in \mathbb{Z}$ such that $$(m-5)^2+(m-4)^2+\ldots+(m+4)^2+(m+5)^2=11(m^2+10)$$ is a perfect square. So, $m \equiv \pm 1 \pmod{11}$ and one can check if $11(m^2+10)$ is a perfect square.

Conjecture:An easy case by case analysis shows that there are no solutions for $11<n \leq 22$. For $n=23$ there is a solution given by $$17^2+18^2+19^2+\ldots+27^2+28^2+29^2=138^2.$$ (Observe that $(m-11)^2+\ldots+(m+11)^2=23(m^2+44)$ and it must be $m \equiv \pm 5 \pmod{23}$ and $m \equiv 0 \pmod{2}$)

If $n>1$ is a natural number $n$ such that there exist $n$ consecutive integers whose sum of their squares is a perfect square, then $n$ is prime.

Mathematical Reflections 2016, Issue 1 - Problem U366

Problem:
If $f:[0,1] \to \mathbb{R}$ is a convex and integrable function with $f(0)=0$, prove that
$$\int_{0}^1 f(x) \ dx \geq 4\int_{0}^{\frac{1}{2}} f(x) \ dx.$$

Proposed by Florin Stanescu, Gaesti, Romania

Solution:
Since $f$ is a convex function, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \int_{0}^1 f(x) \ dx= \displaystyle \dfrac{1}{2}\left[\int_{0}^1 f(x) \ dx+\int_{0}^1 f(1-x) \ dx \right]&=&\displaystyle \int_{0}^1 \dfrac{f(x)+f(1-x)}{2} \ dx \\ &\geq& \displaystyle \int_{0}^1 f\left(\dfrac{x+(1-x)}{2}\right) \ dx\\&=&f\left(\dfrac{1}{2}\right). \end{array}$$
On the other hand, since $f$ is convex and $f(0)=0$, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle f\left(\dfrac{1}{2}\right)=2\cdot\dfrac{f(0)+f\left(\frac{1}{2}\right)}{2}&=& \displaystyle 2\int_{0}^1 \left[(1-x)f(0)+xf\left(\dfrac{1}{2}\right)\right] \ dx \\ & \geq & \displaystyle 2\int_{0}^1 f\left((1-x)\cdot0+x\cdot\dfrac{1}{2}\right) \ dx \\&=& \displaystyle 4\int_{0}^{\frac{1}{2}} f(x) \ dx, \end{array}$$ and the conclusion follows.

Mathematical Reflections 2016, Issue 1 - Problem U365

Problem:
Let $n$ be a positive integer. Evaluate

(a) $\displaystyle \int_{0}^n e^{\lfloor x \rfloor} \ dx$

(b) $\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx$.

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

Solution:
(a) If $k \leq x < k+1$, where $k \in \mathbb{Z}$, then $\lfloor x \rfloor =k$. So, $$\displaystyle \int_{0}^n e^{\lfloor x \rfloor} \ dx=\sum_{k=0}^{n-1} \int_{k}^{k+1} e^{\lfloor x \rfloor} dx=\sum_{k=0}^{n-1} e^k=\dfrac{e^n-1}{e-1}.$$

(b) If $k \leq e^x < k+1$, where $k \in \mathbb{N}^*$, then $\lfloor e^x \rfloor=k$. This implies that if $\log k \leq x < \log (k+1)$, then $\lfloor e^x \rfloor=k$. Let $m$ be the greatest natural number such that $\log m \leq n$.
So, $$\renewcommand{\arraystretch}{2}\begin{array}{lll}\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx&=& \displaystyle \sum_{k=1}^{m-1} \int_{\log k}^{\log k+1} k \ dx+\int_{\log m}^n m \ dx\\&=&\displaystyle \sum_{k=1}^{m-1} [(k+1)\log(k+1)-k\log k]-\sum_{k=1}^{m-1} \log(k+1)+m(n-\log m)\\&=&\displaystyle m\log m-\log m!+mn-m\log m\\&=&mn-\log m!\\&=&\lfloor e^n \rfloor n-\log (\lfloor e^n \rfloor!). \end{array}$$

Mathematical Reflections 2016, Issue 1 - Problem U364

Problem:
Evaluate $$\int \dfrac{5x^2-x-4}{x^5+x^4+1}\ dx.$$

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

Solution:
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{5x^2-x-4}{x^5+x^4+1}&=&\dfrac{5x^2-x-4}{(x^3-x+1)(x^2+x+1)}\\&=&-\dfrac{3(x+1)}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}\\&=&-\dfrac{3}{2}\cdot\dfrac{2x+1}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}-\dfrac{3}{2}\cdot\dfrac{1}{x^2+x+1} \\ &=&-\dfrac{3}{2}\cdot\dfrac{2x+1}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}-\sqrt{3}\cdot\dfrac{2/\sqrt{3}}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1}. \end{array}$$
So, $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \int \dfrac{5x^2-x-4}{x^5+x^4+1}\ dx&=&\displaystyle -\dfrac{3}{2}\int \dfrac{2x+1}{x^2+x+1} \ dx+\int \dfrac{3x^2-1}{x^3-x+1} \ dx-\sqrt{3}\int \dfrac{2/\sqrt{3}}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1} \ dx \\ &=&-\dfrac{3}{2}\log(x^2+x+1)+\log|x^3-x+1|-\sqrt{3}\arctan \dfrac{2x+1}{\sqrt{3}}+C. \end{array}$$

Mathematical Reflections 2016, Issue 1 - Problem S365

Problem:
Let $$a_k=\dfrac{(k^2+1)^2}{k^4+4}, \qquad k=1,2,3,\ldots.$$ Prove that for every positive integer $n$, $$a_1^na_2^{n-1}a_3^{n-2}\cdot \ldots \cdot a_n=\dfrac{2^{n+1}}{n^2+2n+2}.$$

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

Solution:
For any $k \in \mathbb{N}^*$, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} a_k&=&\dfrac{(k^2+1)^2}{(k^2+2)^2-4k^2}\\&=&\dfrac{(k^2+1)^2}{(k^2-2k+2)(k^2+2k+2)}\\&=&\dfrac{k^2+1}{(k-1)^2+1}\cdot\dfrac{k^2+1}{(k+1)^2+1}. \end{array}$$
Let $P_j=\displaystyle \prod_{k=1}^j a_k$, where $j \in \mathbb{N}^*$. Then, $$P_j=\prod_{k=1}^j \dfrac{k^2+1}{(k-1)^2+1} \prod_{k=1}^j \dfrac{k^2+1}{(k+1)^2+1}=(j^2+1)\cdot\dfrac{2}{(j+1)^2+1}.$$ Therefore, we have
$$\renewcommand{\arraystretch}{2} \begin{array}{lll}\displaystyle a_1^na_2^{n-1}a_3^{n-2}\cdot \ldots \cdot a_n&=& \displaystyle \prod_{j=1}^n P_j\\ &=&\displaystyle\prod_{j=1}^n \dfrac{2(j^2+1)}{(j+1)^2+1}\\ &=& \displaystyle 2^n \prod_{j=1}^n \dfrac{j^2+1}{(j+1)^2+1}\\ &=&\displaystyle 2^n\cdot \dfrac{2}{(n+1)^2+1}\\&=&\displaystyle \dfrac{2^{n+1}}{n^2+2n+2}. \end{array}$$

Mathematical Reflections 2016, Issue 1 - Problem S362

Problem:
Let $0 < a,b,c,d \leq 1$. Prove that $$\dfrac{1}{a+b+c+d} \geq \dfrac{1}{4}+\dfrac{64}{27}(1-a)(1-b)(1-c)(1-d).$$

Proposed by An Zhen-ping, Xianyang Normal University, China

Solution:
The given inequality can be written as $$\dfrac{4-a-b-c-d}{4(a+b+c+d)} \geq \dfrac{64}{27}(1-a)(1-b)(1-c)(1-d).$$
By the AM-GM Inequality, we have $$\dfrac{64}{27}(1-a)(1-b)(1-c)(1-d) \leq \dfrac{64}{27}\left(\dfrac{4-a-b-c-d}{4}\right)^4=\dfrac{1}{4}\cdot\dfrac{1}{27}(4-a-b-c-d)^4.$$
So, we have to prove that $$\dfrac{1}{4}\cdot\dfrac{1}{27}(4-a-b-c-d)^4 \leq \dfrac{4-a-b-c-d}{4(a+b+c+d)},$$ i.e. $$\dfrac{1}{27}(4-a-b-c-d)^3 \leq \dfrac{1}{a+b+c+d}.$$ Let $x=a+b+c+d$. We have to prove that $x(4-x)^3 \leq 27$ for any $x \in (0,4]$, but this is true because $$27-x(4-x)^3=(x-1)^2(x^2-10x+27) \geq 0$$ for all $x \in (0,4]$. The equality holds if and only if $x=1$ and $1-a=1-b=1-c=1-d$, i.e. if and only if $a=b=c=d=\dfrac{1}{4}$.

Mathematical Reflections 2016, Issue 1 - Problem S361

Problem:
Find all integers $n$ for which there are integers $a$ and $b$ such that $(a+bi)^4=n+2016i$.

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

Solution:
We have $(a+bi)^4=(a^4-6a^2b^2+b^4)+4ab(a^2-b^2)i$, so it must be
$$\begin{array}{rcl} a^4-6a^2b^2+b^4&=&n \\ ab(a^2-b^2)&=&504. \end{array}$$
If $(a,b)$ is a solution to this system of equations, then also $(-a,-b)$, $(b,-a)$ and $(-b,a)$ are solutions, so we can assume $a^2-b^2>0$ and so $ab>0$. If $a^2-b^2 \equiv 0 \pmod{4}$, then both $a$ and $b$ are even and $ab(a^2-b^2)$ is divisible by $16$, contradiction.  Moreover, since $(a^2-b^2) \ | \ 504$ and $a^2-b^2 \not \equiv 2 \pmod{4}$ for all integers $a,b$, then $(a^2-b^2) \in \{1,3,7,9,21,63\}$. If $a^2-b^2=1$, then $a=\pm 1$ and $b=0$, which gives $ab=0$, contradiction. If $a^2-b^2=3$, then $a=\pm 2$ and $b=\pm 1$, which gives $ab=2$, contradiction. If $a^2-b^2=7$, then $a=\pm 4$ and $b=\pm 3$, which gives $ab=12$, contradiction. If $a^2-b^2=9$, then $a=\pm 5$ and $b=\pm 4$, which gives $ab=20$, contradiction. If $a^2-b^2=21$, then $a=\pm 11$ and $b=\pm 10$ or $a=\pm 5$ and $b=\pm 2$, which gives $ab \in \{110,10\}$, contradiction. If $a^2-b^2=63$, then $a=\pm 32$ and $b=\pm 31$ or $a=\pm 12$ and $b=\pm 9$ or $a=\pm 8$ and $b=\pm 1$, which gives $ab \in \{992,108,8\}$. We conclude that $a^2-b^2=63$ and $ab=8$, so $$a^4-6a^2b^2+b^4=(a^2-b^2)^2-4(ab)^2=3713,$$ which yields $n=3713$.

Mathematical Reflections 2016, Issue 1 - Problem J363

Problem:
Solve in integers the system of equations $$\begin{array}{rcl} x^2+y^2-z(x+y)&=&10 \\ y^2+z^2-x(y+z)&=&6 \\ z^2+x^2-y(z+x)&=&-2. \end{array}$$

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

Solution:

Observe that if $(x_0,y_0,z_0)$ is a solution to the given system, then also $(-x_0,-y_0,-z_0)$ is a solution. So, if there exists a solution, we can assume without loss of generality that $x_0+y_0+z_0 \geq 0$.
Adding the first equation to the third equation and subtracting the second, we get $$x^2-yz=1.$$ Adding the first equation to the second equation and subtracting the third, we get $$y^2-zx=9.$$ Adding the second equation to the third equation and subtracting the first, we get $$z^2-xy=-3.$$ So, the given system becomes:
$$\begin{array}{rcl} x^2-yz&=&1 \\ y^2-zx&=&9 \\ z^2-xy&=&-3. \end{array}$$
Subctracting the first two equations, then the last two equations and then the first equation and the third equation, we obtain the system of equations
$$\begin{array}{rcl} (x-y)(x+y+z)&=&-8 \\ (y-z)(x+y+z)&=&12 \\ (z-x)(x+y+z)&=&-4. \end{array}$$
Since by our assumption we have $x+y+z \geq 0$, we obtain that $x+y+z \in \{1,2,4\}$. We have three cases.

(i) $x+y+z=1$. Hence, we have $x-y=-8$ and $z-x=-4$, which gives $y=x+8$ and $z=x-4$. Therefore, $x+(x+8)+(x-4)=1$, which gives $x=-1,y=7,z=-5$.
(ii) $x+y+z=2$. Hence, we have $x-y=-4$ and $z-x=-2$, which gives $y=x+4$ and $z=x-2$. Therefore, $x+(x+4)+(x-2)=2$, which gives $x=0,y=4,z=-2$.
(iii) $x+y+z=4$. Hence, we have $x-y=-2$ and $z-x=-1$, which gives $y=x+2$ and $z=x-1$. Therefore, $x+(x+2)+(x-1)=4$, which gives $x=1,y=3,z=0$.

An easy check shows that the only solutions are $$(x,y,z) \in \{(1,3,0),(-1,-3,0)\}.$$

Mathematical Reflections 2016, Issue 1 - Problem J362

Problem:
Let $a,b,c,d$ be real numbers such that $abcd=1$. Prove that the following inequality holds: $$ab+bc+cd+da \leq \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2}.$$

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution:
Without loss of generality, assume that $a \leq b \leq c \leq d$. Then, $\dfrac{1}{a} \geq \dfrac{1}{b} \geq \dfrac{1}{c} \geq \dfrac{1}{d}$. By the Rearrangement Inequality, $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2} \geq \dfrac{1}{cd}+\dfrac{1}{da}+\dfrac{1}{ab}+\dfrac{1}{bc}=ab+bc+cd+da.$$

Mathematical Reflections 2016, Issue 1 - Problem J361

Problem:
Solve in positive integers the equation $$\dfrac{x^2-y}{8x-y^2}=\dfrac{y}{x}.$$

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

Solution:

The given equation can be retwritten as $$x^3+y^3+27-9xy=27.$$ Using the well-known identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$ we obtain $$(x+y+3)(x^2+y^2+9-xy-3x-3y)=27.$$ Since $x,y$ are positive integers, then $x+y+3 \geq 5$ and we have two cases.

(i) $x+y+3=9$ and $x^2+y^2+9-xy-3(x+y)=3$. From the first equation we have $x+y=6$, which gives $x^2+y^2=36-2xy$. Substituting these values into the second equation, we have $(36-2xy)+9-xy-18=3$, which gives $xy=8$. So, we obtain the system of equations
$$\begin{array}{rcl} x+y&=&6 \\ xy&=&8, \end{array}$$ which gives $(x,y) \in \{(2,4),(4,2)\}$. An easy check shows that the only solution to the given equation is $(x,y)=(4,2)$.

(ii) $x+y+3=27$ and $x^2+y^2+9-xy-3x-3y=1$.  From the first equation we have $x+y=24$, which gives $x^2+y^2=576-2xy$. Substituting these values into the second equation, we have $(576-2xy)+9-xy-72=3$, which gives $xy=170$. So, we obtain the system of equations $$\begin{array}{rcl} x+y&=&24 \\ xy&=&170, \end{array}$$ which yields no solutions in positive integers.

In conclusion, $(x,y)=(4,2)$. 

Gazeta Matematica 11/2015, Problem E:14911

Problem:
In a school, each student knows exactly $2n+1$ other students, where $n \in \mathbb{N}^*$.
Prove that the number of students in the school is even. (Suppose that "`knowing"' is a symmetrical relation).

Proposed by Alessandro Ventullo, Milan, Italy

Solution:

Suppose that two students know each other if and only if they shake their hands. Let $N$ be the total number of the handshakings and let $m$ be the total number of students. Since each student shakes his hand with $2n+1$ other students, there are $m(2n+1)$ handshakings. Since in each handshake are involved two students, we have counted each handshaking twice (one time when student $A$ shakes his hand with student $B$ and one time when student $B$ shakes his hand with student $A$). Therefore, the total number of handshakings is $N=\dfrac{m(2n+1)}{2}$. Since $N$ is a natural number, then $m$ must be even. 

Gazeta Matematica 10/2015, Problem 27137

Problem:
Evaluate $$\lim_{y \to 0} \int_{y}^1 \left(\arctan x \arctan \dfrac{1}{x}+\dfrac{\log(1+x^2)}{1+x^2}\right) \ dx.$$

Proposed by Alessandro Ventullo, Milan, Italy

 

Solution:

Observe that $$\arctan x + \arctan \dfrac{1}{x}=\dfrac{\pi}{2}$$ for all $x>0$. Setting $a=\arctan x, b=\arctan \dfrac{1}{x}, c=-\dfrac{\pi}{2}$, we have $a+b+c=0$, so $$a^3+b^3+c^3=3abc.$$ This means that
$$-3\arctan x \cdot \arctan \dfrac{1}{x} \cdot \dfrac{\pi}{2}=\arctan^3 x + \arctan^3 \dfrac{1}{x}-\dfrac{\pi^3}{8}.$$
Hence
$$
\int_0^1 \arctan x \arctan \dfrac{1}{x} \ dx=-\dfrac{2}{3\pi}\int_0^1 \left( \arctan^3 x + \arctan^3 \dfrac{1}{x}-\dfrac{\pi^3}{8}\right) \ dx.                  (1)$$
Integrating by part, we obtain
$$
\int \arctan^3 x \ dx=x \arctan^3 x - \dfrac{3}{2}\log(1+x^2)\arctan^2 x+3 \int \dfrac{\log (1+x^2) \arctan x}{1+x^2} \ dx.
$$
and
$$
\int \arctan^3 \dfrac{1}{x} \ dx=x \arctan^3 \dfrac{1}{x} + \dfrac{3}{2}\log(1+x^2)\arctan^2 \dfrac{1}{x}+3 \int \dfrac{\log (1+x^2) \arctan 1/x}{1+x^2} \ dx.
$$
Summing up and integrating from $x=0$ to $x=1$ and using the fact that $$\arctan x+\arctan \dfrac{1}{x}=\dfrac{\pi}{2},$$ we get
$$
\int_0^1 \left( \arctan^3 x + \arctan^3 \dfrac{1}{x} \right) \ dx=\dfrac{\pi^3}{32}+\dfrac{3\pi}{2} \int_0^1 \dfrac{\log (1+x^2)}{1+x^2} \ dx.                 (2)
$$
Substituting (1) into (2), we get
$$\int_{0}^1 \left(\arctan x \arctan \dfrac{1}{x}+\dfrac{\log(1+x^2)}{1+x^2}\right)=\dfrac{\pi^2}{16}.$$