Mathematical Reflections 2014, Issue 1 - Problem O289

Problem:
Let $a, b, x, y$ be positive real numbers such that $x^2-x+1=a^2$, $y^2+y+1=b^2$, and
$(2x-1)(2y+1)=2ab+3$. Prove that $x+y=ab$.

Proposed by Titu Andreescu.

Solution:
Multiplying both sides of the first two equations by $4$ and both sides of the third equation by $2$ , we have
$$
\begin{array}{rll}
(2x-1)^2+3&=&4a^2            (1)\\
(2y+1)^2+3&=&4b^2            (2)\\
2(2x-1)(2y+1)&=&4ab+6     (3)\\
\end{array}
$$
Summing up these equations, we get $$(2x-1)^2+2(2x-1)(2y+1)+(2y+1)^2+6=4(a^2+ab+b^2)+6,$$ i.e. $$[(2x-1)+(2y+1)]^2=4(a^2+ab+b^2),$$ which gives $x+y=\sqrt{a^2+ab+b^2}$. So, it suffices to show that $\sqrt{a^2+ab+b^2}=ab$. Multiplying $(1)$ and $(2)$, we get $[(2x-1)^2+3][(2y+1)^2+3]=16a^2b^2$, and using $(3)$ we get $$[(2ab+3)^2+3(4a^2-3+4b^2-3)+9]=16a^2b^2,$$ which gives $a^2+ab+b^2=a^2b^2$. Taking the square root, we obtain the conclusion.

Mathematical Reflections 2014, Issue 1 - Problem U294

Problem:
Let $p_1, p_2, \ldots, p_n$ be pairwise distinct prime numbers. Prove that
$$\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})=\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n}).$$

Proposed by Marius Cavachi.

Solution:
Clearly, $\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n})$ is a subfield of $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$. Observe that $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ is Galois over $\mathbb{Q}$, since it is the splitting field of the polynomial $(x^2-\sqrt{p_1})\cdots(x^2-\sqrt{p_n})$. Every automorphism $\sigma$ is completely determined by its action on $\sqrt{p_1},\ldots,\sqrt{p_n}$, which must be mapped to $\pm \sqrt{p_1},\ldots,\pm \sqrt{p_n}$, respectively. Therefore, $\textrm{Gal}(\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})/\mathbb{Q})$ is the group generated by $\{\sigma_1,\sigma_2,\ldots,\sigma_n\}$, where $\sigma_i$ is the automorphism defined by $$\sigma_i(\sqrt{p_j})=\begin{cases} -\sqrt{p_j} & \textrm{if } i=j \\ \sqrt{p_j} & \textrm{if } i \neq j. \end{cases}$$ Clearly, the only automorphism that fixes $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ is the identity. Moreover, it's easy to see that the only automorphism that fixes the element $\sqrt{p_1}+\ldots+\sqrt{p_n}$ is the identity, which means that the only automorphism that fixes $\mathbb{Q}(\sqrt{p_1}+\ldots+\sqrt{p_n})$ is the identity. Hence, by the Fundamental Theorem of Galois Theory, $\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n})=\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$.

Mathematical Reflections 2014, Issue 1 - Problem U289

Problem:
Let $a \geq 1$ be such that $\left(\left\lfloor a^n \right\rfloor \right)^{\frac{1}{n}} \in \mathbb{Z}$ for all sufficiently large integers $n$. Prove that $a \in \mathbb{Z}$.

Proposed by Mihai Piticari.

Solution:
Suppose that $\left(\lfloor a^n \rfloor \right)^{\frac{1}{n}}=m_n$ for all $n \geq n_0$, where $n_0 \in \mathbb{Z}^+$ and $m_n \in \mathbb{Z}$. Hence, $\lfloor a^n \rfloor=m_n^n$, so for all $n \geq n_0$ we have $a^n-1 < m_n^n \leq a^n$, i.e. $$\sqrt[n]{a^n-1} < m_n \leq a \qquad \forall n \geq n_0.$$
By the Squeeze Theorem, since $\displaystyle \lim_{n \to \infty} \sqrt[n]{a^n-1}=\lim_{n \to \infty} a=a$, we have that $\displaystyle \lim_{n \to \infty} m_n=a$. Since $\{m_n\}_{n \geq n_0}$ is a convergent sequence of integers, therefore must be constant for all $n \geq n_1 \geq n_0$. Indeed, $\displaystyle \lim_{n \to \infty} m_n=a$ implies that there exists $n_1 \in \mathbb{Z}^+$ such that for all $n \geq n_1$ we have $a-1/2<m_n<a+1/2$. Since $(a-1/2,a+1/2)$ has length $1$, it follows that there is only $m_n$ in this interval. This means that $m_n$ is constant for all $n \geq n_1$ and so $m_n=a$ for all $n \geq n_1$, which gives $a \in \mathbb{Z}$.

Mathematical Reflections 2014, Issue 1 - Problem S294

Problem:
Let $s(n)$ be the sum of digits of $n^2+1$. Define the sequence $(a_n)_{n \geq 0}$ by $a_{n+1} = s(a_n)$, with $a_0$ an arbitrary positive integer. Prove that there is $n_0$ such that $a_{n+3} = a_n$ for all $n \geq n_0$.

Proposed by Roberto Bosch Cabrera.

Solution:
We have to prove that the given sequence is $3$-periodic. Since $f(5)=8, f(8)=11$ and $f(11)=5$, it suffices to prove that for every positive integer $a_0$ there exists some $n \in \mathbb{N}$ such that $a_n \in \{5,8,11\}$. Let $m$ be the number of digits of $a_0$. We prove the statement by induction on $m$. For $m \leq 2$ we proceed by a direct check. If $a_0 \in \{5,8,11\}$ there is nothing to prove. If $a_0$ is a two-digit number, then $a^2_0 \leq 10000$, so $a_1 \leq 37$ and we reduce to analyze the cases for $a_0 \leq 37$.

(i) If $a_0 \in \{2,7,20\}$, then $a_1=5$. If $a_0 \in \{1,10,26,28\}$, then $a_1 \in \{2,20\}$, so $a_2=5$. Finally, if $a_0 \in \{3,6,9,12,15,18,27,30,33\}$, then $a_3=5$.
(ii) If $a_0 \in \{4,13,23,32\}$, then $a_1=8$.
(iii) If $a_0 \in \{17,19,21,35,37\}$, then $a_1=11$. If $a_0 \in \{14,22,24,31,36\}$, then $a_1 \in \{17,19\}$, so $a_2=11$. Finally, if $a_0 \in \{16,25,29,34\}$, then $a_3=11$.

Thus, we have proved that if $a_0$ is a one or two-digit number, then $a_n \in \{5,8,11\}$ for some $n \in \mathbb{N}$, i.e. the sequence is $3$-periodic. Let $m \geq 2$ and suppose that the statement is true for all $k \leq m$. Let $a_0$ be an $(m+1)$-digit number. Then, $10^m \leq a_0 < 10^{m+1}$ which implies $10^{2m} \leq a^2_0 < 10^{2(m+1)}$. Hence, $$a_1=f(a_0) \leq 9\cdot2(m+1)+1<10^m,$$ and by induction hypothesis, the sequence $(a_n)_{n \geq 1}$ is $3$-periodic, which implies that the sequence $(a_n)_{n \geq 0}$ is $3$-periodic, as we wanted to prove. 

Mathematical Reflections 2014, Issue 1 - Problem S290

Problem:
Prove that there is no integer $n$ for which $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{n^2}=\left(\dfrac{4}{5}\right)^2.$$

Proposed by Ivan Borsenco.

Solution:
Let $p$ be the greatest prime number such that $p \leq n < 2p$. Then the given equality can be written as $$5^2k=(4\cdot n!)^2,$$ where $k=\displaystyle \sum_{i=1}^n \dfrac{(n!)^2}{i^2}$. Observe that $k \equiv \dfrac{(n!)^2}{p^2} \not \equiv 0 \pmod{p}$. Since $p|5^2k$ and $p$ does not divide $k$, it follows that $p|5^2$, i.e. $p=5$. So, $n \in \{5,6,7,8,9\}$. An easy check shows that none of these values satisfies the equality.

Mathematical Reflections 2014, Issue 1 - Problem J293

Problem:
Find all positive integers $x, y, z$ such that $$(x+y^2+z^2)^2-8xyz=1.$$

Proposed by Aaron Doman.

Solution:
We rewrite the equation as $$x^2+2x(y^2+z^2-4yz)+(y^2+z^2)^2-1=0.$$ Since $x$ must be a positive integer, the discriminant of this quadratic equation in $x$ must be non-negative, i.e. $$(y^2+z^2-4yz)^2-(y^2+z^2)^2+1 \geq 0,$$ which is equivalent to $$-8yz(y-z)^2+1 \geq 0,$$ which gives $yz(y-z)^2 \leq 1/8$. Since $y$ and $z$ are positive integers, it follows that $$yz(y-z)^2=0,$$ so $y-z=0$, i.e. $y=z$. The given equation becomes $(x-2y^2)^2=1$, which yields $x=2y^2 \pm 1$. Therefore, all the positive integer solutions to the given equation are $$(2n^2-1,n,n), (2n^2+1,n,n), \qquad n \in \mathbb{Z}^+.$$

Mathematical Reflections 2014, Issue 1 - Problem J292

Problem:
Find the least real number $k$ such that for every positive real numbers $x, y, z$, the following
inequality holds: $$\prod_{cyc} (2xy+yz+zx) \leq k(x+y+z)^6.$$

Proposed by Dorin Andrica.

Solution:
Observe that $$(x-y)^2+z^2 \geq 0 \iff x^2+y^2+z^2-2xy \geq 0 \iff (x+y+z)^2 \geq 4xy+2yz+2zx,$$ so $2xy+yz+zx \leq \dfrac{1}{2}(x+y+z)^2$. Likewise, $2yz+zx+xy \leq \dfrac{1}{2}(x+y+z)^2$ and $2zx+xy+yz \leq \dfrac{1}{2}(x+y+z)^2$. Therefore,
$$\prod_{cyc} (2xy+yz+zx) \leq \dfrac{1}{8}(x+y+z)^6.$$ The equality is attained if and only if $(x-y)^2+z^2=0, (y-z)^2+x^2=0, (z-x)^2+y^2=0$, i.e. if and only if $x=y=z=0$. Clearly, $k_{\min}=1/8$, because for each inequality $1/2$ was the least real number for which the inequality was satisfied.


Note. Actually, this is not the minimum value of $k$, which is $64/729$. See the official solution.

Mathematical Reflections 2014, Issue 1 - Problem J290

Problem:
Let $a, b, c$ be nonnegative real numbers such that $a + b + c = 1$. Prove that
$$\sqrt[3]{13a^3+14b^3}+\sqrt[3]{13b^3+14c^3}+\sqrt[3]{13c^3+14a^3} \geq 3.$$

Proposed by Titu Andreescu.

Solution:

The second Minkowski's Inequality (see \emph{Zdravko Cvetkovski - Inequalities. Theorems, Techniques and Selected Problems, page 99}) states that if $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ are positive real numbers and $p>1$, then
$$\left(\left(\sum_{i=1}^n a_i\right)^p+\left(\sum_{i=1}^n b_i\right)^p \right)^{1/p} \leq \sum_{i=1}^n (a^p_i+b^p_i)^{1/p}.$$
Equality occurs if and only if $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\ldots=\dfrac{a_n}{b_n}$.\\
Now, take $$(a_1,a_2,a_3)=(\sqrt[3]{13}a,\sqrt[3]{13}b,\sqrt[3]{13}c),$$ $$(b_1,b_2,b_3)=(\sqrt[3]{14}b, \sqrt[3]{14}c,\sqrt[3]{14}a)$$ and $p=3$. Then,
$$\begin{array}{lll} \sqrt[3]{13a^3+14b^3}+\sqrt[3]{13b^3+14c^3}+\sqrt[3]{13c^3+14a^3} & \geq & \sqrt[3]{\left(\sqrt[3]{13(a+b+c)}\right)^3+\left(\sqrt[3]{14(a+b+c)}\right)^3} \\ & = & \sqrt[3]{(\sqrt[3]{13})^3+(\sqrt[3]{14})^3} \\ &=& 3. \end{array}$$
Equality occurs if and only if $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}$. 

Mathematical Reflections 2014, Issue 1 - Problem J289

Problem:
Let $a$ be a real number such that $0 \leq a < 1$. Prove that
$$\left\lfloor a\left(1+\left\lfloor \dfrac{1}{1-a} \right\rfloor \right)\right\rfloor+1=\left\lfloor \dfrac{1}{1-a} \right\rfloor.$$

Proposed by Arkady Alt.

Solution:

Let $m \in \mathbb{N}$ and $\alpha \in [0,1)$ such that $\dfrac{1}{1-a}=m+\alpha$. Then, we have to prove that
$$\left\lfloor \left(1-\dfrac{1}{m+\alpha}\right)(1+m) \right\rfloor+1=m.$$
Indeed, $$\left\lfloor \left(1-\dfrac{1}{m+\alpha}\right)(1+m) \right\rfloor=\left\lfloor m+1-\dfrac{m+1}{m+\alpha} \right\rfloor=m-1,$$ where the last equality follows from $-2<-\dfrac{m+1}{m+\alpha}<-1$.