Gazeta Matematica 12/2015, Problem S:E15.354

Problem:
Prove that the number $$a=2n^2+[\sqrt{4n^2+n}]+1$$ where $n$ is a nonzero natural number, can be written as the sum of two perfect squares.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let us prove first that $[\sqrt{4n^2+n}]=2n$. Indeed, for all $n \in \mathbb{N}$ $$4n^2 \leq 4n^2+n < 4n^2+4n+1,$$ so $$2n \leq \sqrt{4n^2+n}<2n+1.$$ It follows that $[\sqrt{4n^2+n}]=2n$. So, $$a=2n^2+2n+1=n^2+(n+1)^2.$$ 

Mathematical Reflections 2016, Issue 2 - Problem U372

Problem:
Let $\alpha,\beta>0$ be real numbers and let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) \neq 0$ for all $x$ in a neighborhood $U$ of $0$. Evaluate $$\lim_{x \to 0^+} \dfrac{\int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\int_0^{\beta x} t^{\beta}f(t) \ dt}.$$

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

Solution:
Observe that $\lim_{x \to 0+} \int_0^{\alpha x} t^{\alpha}f(t) \ dt=\lim_{x \to 0^+} \int_0^{\beta x} t^{\beta}f(t) \ dt=0$. Let $g(t)=t^{\alpha}f(t)$ and $h(t)=t^{\beta} f(t)$. We have $$\dfrac{d}{dx} \int_0^{\alpha x} g(t) \ dt=g(\alpha x)\alpha=(\alpha x)^{\alpha}f(\alpha x)\alpha$$ and $$\dfrac{d}{dx} \int_0^{\beta x} h(t) \ dt=h(\beta x)\beta=(\beta x)^{\beta}f(\beta x)\beta.$$ Clearly, $\dfrac{d}{dx} \int_0^{\beta x} h(t) \ dt \neq 0$ for all $x$ in a neighborhood $U$ of $0$ ($x \neq 0$). We have $$\lim_{x \to 0^+}\dfrac{\dfrac{d}{dx} \int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\dfrac{d}{dx} \int_0^{\beta x} t^{\beta}f(t) \ dt}=\begin{cases} 0 & \textrm{ if } \alpha>\beta \\ 1 & \textrm{ if } \alpha=\beta \\ +\infty & \textrm{ if } \alpha<\beta.  \end{cases}$$
By L'H\^{o}pital's Rule we conclude that
$$\lim_{x \to 0^+} \dfrac{\int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\int_0^{\beta x} t^{\beta}f(t) \ dt}=\begin{cases} 0 & \textrm{ if } \alpha>\beta \\ 1 & \textrm{ if } \alpha=\beta \\ +\infty & \textrm{ if } \alpha<\beta.  \end{cases}$$

Mathematical Reflections 2016, Issue 2 - Problem U370

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1}{x}.$$

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

Solution:
Observe that for any $k=2,\ldots,n$ and for any $x$ such that $|x|<1$, we have $$\sqrt[k]{1+kx}=1+\dfrac{1}{k}kx+o(x^2)=1+x+o(x^2).$$ Hence,
$$\begin{array}{lll}\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1&=&(1+x+o(x^2))^{n-1}-1\\&=&(1+(n-1)x+o(x^2))-1\\&=&(n-1)x+o(x^2). \end{array}$$
Therefore, $$\lim_{x \to 0} \dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1}{x}=n-1.$$

Mathematical Reflections 2016, Issue 2 - Problem U368

Problem:
Let $$x_n=\sqrt{2}+\sqrt[3]{\dfrac{3}{2}}+\ldots+\sqrt[n+1]{\dfrac{n+1}{n}}, \qquad n=1,2,3,\ldots$$
Evaluate $$\lim_{n \to \infty} \dfrac{x_n}{n}.$$

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

Solution:

Let $y_n=n$. Since $(y_n)_{n \geq 1}$ is strictly monotone and divergent sequence and
$$\lim_{n \to \infty} \dfrac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n \to \infty} \left(1+\dfrac{1}{n+1}\right)^{\frac{1}{n+2}}=1,$$ then by the Stolz-Cesaro Theorem we have $$\lim_{n \to \infty} \dfrac{x_n}{n}=1.$$ 

Mathematical Reflections 2016, Issue 2 - Problem S368

Problem:
Determine all the natural numbers $n$ such that $$\sigma(n)=n+55,$$ where $\sigma(n)$ denotes the sum of the divisors of $n$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $\sigma(n) \geq n+1$, then the sum of the proper divisors of $n$ is $54$. Denoting by $\omega(n)$ the number of distinct primes dividing $n$, observe that if $\omega(n) \geq 4$, then there exist four prime numbers $p<q<r<s$ that divide $n$. Since $p \geq 2$, $q \geq 3$, $r \geq 5$ and $s \geq 7$, then $qrs$ divides $n$ but $qrs \geq 3\cdot5\cdot7>54$, which contradicts the hypothesis. So, $\omega(n) \leq 3$. We have three cases.

(i) $\omega(n)=1$. Then, $n=p^k$, where $p$ is a prime number and $k \in \mathbb{N}^*$. Hence, $$1+p+\ldots+p^k=p^k+55 \implies p(1+p+\ldots+p^{k-2})=54.$$ It follows that $p \in \{2,3\}$. An easy check shows that there are no solutions.

(ii) $\omega(n)=2$. Then, $n=p^{k_1}q^{k_2}$, where $p,q$ are prime numbers, $p<q$ and $k_1,k_2 \in \mathbb{N}^*$. Since $p \geq 2$ and $q \geq 3$, if $n=p^4q$, then $p+p^2+p^3+p^4+p^3q+q \geq 57>54$, so $k_1 \leq 3$. If $n=pq^3$, then $pq^2+q+q^2+q^3 \geq 2\cdot9+39>54$, so $k_2 \leq 2$.  We have six cases.

(a) $k_1=1,k_2=1$. Then, $n=pq$ and $p+q=54$, which gives $$(p,q) \in \{(7,47),(11,43),(13,41),(17,37),(23,31)\}.$$

(b) $k_2=2, k_2=1$. Then, $n=p^2q$ and $p+p^2+q+pq=54$. Both $p$ and $q$ must be odd. Since $p+p^2<54$, then $p \in \{3,5\}$. An easy check shows that there are no solutions.

(c) $k_1=3, k_2=1$. Then, $n=p^3q$ and $p+p^2+p^3+q+pq+p^2q=54$. Both $p$ and $q$ must be odd. Since $5^3>54$, then $p=3$. An easy check shows that there are no solutions.

(d) $k_1=1, k_2=2$. Then, $n=pq^2$ and $p+pq+q+q^2=54$. Since $q+q^2<54$, then $q \in \{3,5\}$. An easy check shows that there are no solutions.

(e) $k_1=2,k_2=2$. Then, $n=p^2q^2$ and $p+p^2+pq+p^2q+pq^2+q+q^2=54$. At least one between $p$ and $q$ must be even, so $p=2$. Thus, we get $$(p,q)=(2,3).$$

(f) $k_1=3, k_2=2$. Then, $n=p^3q^2$, but $p^3q+p^2q^2 \geq 8\cdot3+4\cdot9>54$, so there are no solutions.

(iii) $\omega(n)=3$. Then, $n=p^{k_1}q^{k_2}r^{k_3}$, where $p,q,r$ are prime numbers, $p<q<r$ and $k_1,k_2,k_3 \in \mathbb{N}^*$. Since $p \geq 2$, $q \geq 3$ and $r \geq 5$, if $n=p^3qr$, then $p^3q+p^3r \geq 8\cdot3+8\cdot5>54$, so $k_1 \leq 2$. If $n=pq^2r$, then $pq^2+q^2r \geq 2\cdot9+9\cdot5>54$, so $k_2=1$. If $n=pqr^2$, then $qr^2 \geq 3\cdot 25>54$, so $k_3=1$.  We have two cases.

(a) $k_1=k_2=k_3=1$. Then, $n=pqr$ and $p+q+r+pq+pr+qr=54$. Clearly, $p \neq 2$. If $p \geq 5$, then $q \geq 7$ and $r \geq 11$, but $pr=55>54$, contradiction. So, $p=3$ and $4q+4r+qr=51$, i.e. $(q+4)(r+4)=67$, but $67$ is a prime number, so no solutions in this case.

(b) $k_2=2, k_2=k_3=1$. Then, $n=p^2qr$ and $p+p^2+q+pq+p^2q+r+pr+p^2r+qr+pqr=54$. Clearly, $p$ cannot be even. So, $p \geq 3$ but $p^2r \geq 9\cdot7>54$, contradiction.


In conclusion, $n \in \{36,329,473,533,629,713\}$.

Mathematical Reflections 2016, Issue 2 - Problem J371

Problem:
Prove that for all positive integers $n$, $${n+3 \choose 2} +6{n+4 \choose 4} + 90{n+5 \choose 6}$$ is the sum of two perfect cubes.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:

Let $$N=\dfrac{(n+5)!}{8(n-1)!}+\dfrac{(n+4)!}{4n!}+\dfrac{(n+3)!}{2(n+1)!}.$$
We have $$N=\dfrac{1}{8}[n(n+1)(n+2)(n+3)(n+4)(n+5)+2(n+1)(n+2)(n+3)(n+4)+4(n+2)(n+3)]$$
Set $x=n^2+5n$. Hence, $$\begin{array}{lll} N&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+6)]\\&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+4)+8]\\&=&\dfrac{1}{8}(x+4)(x(x+6)+2(x+6)+4)+1\\&=&\dfrac{1}{8}(x+4)^3+1\\&=&\left(\dfrac{n^2+5n+4}{2}\right)^3+1\\&=&\left(\dfrac{(n+1)(n+4)}{2}\right)^3+1,\end{array}$$ which is the sum of two perfect cubes since $(n+1)(n+4)$ is even for all $n \in \mathbb{N}^*$.

Mathematical Reflections 2016, Issue 2 - Problem J369

Problem:
Solve the equation $$\sqrt{1+\dfrac{1}{x+1}}+\dfrac{1}{\sqrt{x+1}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}.$$

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

Solution:
Clearly, $x>0$. The given equation can be written as $$\dfrac{\sqrt{x+2}}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}}=\dfrac{x+1}{\sqrt{x}},$$ i.e. $$\sqrt{x^2+2x}+\sqrt{x}=(x+1)\sqrt{x+1}.$$ The last equation can be written as $$\dfrac{\sqrt{x}}{\sqrt{x+2}-1}=\sqrt{x+1},$$ which gives $$\sqrt{x}+\sqrt{x+1}=\sqrt{(x+2)(x+1)}.$$ Squaring both sides and reordering, we get $$x^2+x+1=2\sqrt{x^2+x},$$ i.e. $$(\sqrt{x^2+x}-1)^2=0.$$ The last equation is equivalent to $x^2+x-1=0$, which solved for $x>0$ gives $x=\dfrac{-1+\sqrt{5}}{2}$.

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, $$\renewcommand{\arraystretch}{2}\begin{array}{lll}\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx&=& \displaystyle \sum_{k=1}^{m-1} \int_{\log k}^{\log k+1} k \ dx+\int_{\log m}^n m \ dx\\&=&\displaystyle \sum_{k=1}^{m-1} [(k+1)\log(k+1)-k\log k]-\sum_{k=1}^{m-1} \log(k+1)+m(n-\log m)\\&=&\displaystyle m\log m-\log m!+mn-m\log m\\&=&mn-\log m!\\&=&\lfloor e^n \rfloor n-\log (\lfloor e^n \rfloor!). \end{array}$$