Mathematical Reflections 2012, Issue 1 - Problem U217

Problem:
Define an increasing sequence $(a_k)_{k \in \mathbb{Z}^+}$ to be attractive if $\sum_{k=1}^\infty \dfrac{1}{a_k}$ diverges and $\sum_{k=1}^\infty \dfrac{1}{a^2_k}$ converges. Prove that there is an attractive sequence $a_k$ such that $a_k\sqrt{k}$ is also an attractive sequence.

Proposed by Ivan Borsenco.

Solution:

We claim that $a_k = \sqrt{k} \log (k+1)$ do the trick. Indeed, $a_k = \sqrt{k} \log (k+1)$ is obviously increasing for all $k \geq 1$. Moreover, since $\dfrac{1}{k} \leq \dfrac{1}{\sqrt{k}\log (k+1)}$ for all $k \geq 1$ and $\displaystyle \int_1^\infty \dfrac{1}{x\log^2 (x+1)} \ dx$ converges we obtain that
$$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k}\log (k+1)} = \infty, \qquad \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k \log^2 (k+1)} = M < \infty$$
as a consequence of the Comparison Test and the Integral Test.
Finally, $a_k\sqrt{k}=k \log (k+1)$ is also attractive. As a matter of fact, $k \log (k+1)$ is increasing for all $k \geq 1$, $\displaystyle \int_1^\infty \dfrac{1}{x\log (x+1)} \ dx$ diverges and $\dfrac{1}{k^2 \log^2 (k+1)} \leq \dfrac{1}{k^2}$ for all $k \geq 2$, therefore
$$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k\log (k+1)} = \infty, \qquad \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k^2 \log^2 (k+1)} = N < \infty.$$

Mathematical Reflections 2012, Issue 1 - Problem J221

Problem:
Solve in integers the system of equations

$\begin{array}{lll} xy-\dfrac{z}{3} = xyz+1 \\ yz-\dfrac{x}{3} = xyz - 1 \\ zx-\dfrac{y}{3} = xyz-9. \end{array}$

Proposed by Titu Andreescu.

Solution:

It's easy to see that $x,y,z$ must be divisible by $3$. Summing up all the equations and factorizing, we obtain $$(x-1)(y-1)(z-1)(xy+yz+zx+1)=8.$$ Then $x-1=\pm1,\pm 2, \pm 4, \pm 8$, i.e. $x=0,2,-1,3,-3,5,-7,9$ and for what we said, we reduce to $x \in \{-3,0,3,9\}$. Now, consider the second equation of the system $$yz-\dfrac{x}{3} = xyz - 1.$$ We have four cases.

(i) If $x=-3$, then $2yz=-1$, so no integer solution.
(ii) If $x=0$, then $yz=-1$, but the first equation gives $z=-3$, so no integer solution.

(iii) If $x=3$, then $yz=0$. If $y=0$ we get $z=-3$ from the first equation of the system and these values satisfy also the third equation, i.e. $(3,0,-3)$ is a solution. If $z=0$, the first equation gives $3y=1$, i.e. no integer solution.

(iv) If $x=9$, then $4yz=-1$, so no integer solution.

Therefore, $(3,0,-3)$ is the only integer solution to the given system.

Mathematical Reflections 2012, Issue 1 - Problem J220

Problem:
Find the least prime $p$ for which $p=a^2_k+kb^2_k, k=1,\ldots,5$, for some $(a_k,b_k)$ in $\mathbb{Z} \times \mathbb{Z}$.

Proposed by Cosmin Pohoata.

Solution:

We have $p \equiv a^2_k \pmod{k}$ for $k=2,\ldots,5$, so $$\begin{array}{lll} p \equiv 1 \pmod{2} \\ p \equiv 1 \pmod{3} \\ p \equiv 1 \pmod{4} \\ p \equiv 1,4 \pmod{5}. \end{array}$$ Then $p=1+60n$ or $p=49+60n$ where $n \in \mathbb{N}$. Clearly $p \neq 1,49$. Moreover, $p \neq 61,109,121,169,181,229$ since $121,169$ are not primes and the equation
$$a^2_2+2b^2_2 \equiv 5 \pmod{8}$$ has no solution. We conclude that $p=241$ is the least prime which satisfies the given conditions, since
$$241=15^2+4^2=13^2+2\cdot6^2=7^2+3\cdot8^2=15^2+4\cdot2^2=14^2+5\cdot3^2.$$ 

Mathematical Reflections 2012, Issue 1 - Problem J219

Problem:
Trying to solve a problem, Jimmy used the following "`formula"': $\log_{ab} x = \log_a x \log_b x$,
where $a, b, x$ are positive real numbers different from $1$. Prove that this is correct only
if $x$ is a solution to the equation $\log_a x + \log_b x = 1$.

Proposed by Titu Andreescu.

Solution:

We have $\log_{ab} x = \dfrac{\log_a x}{1+\log_a b}$. So $$\log_{ab} x = \log_a x \log_b x \iff \dfrac{1}{1+\log_a b}=\log_b x \iff 1=\log_b x + \log_a b \log_b x.$$ Since $\log_b x = \dfrac{\log_a x}{\log_a b}$, we are done.

Mathematical Reflections 2012, Issue 1 - Problem J217

Problem:
If $a, b, c$ are integers such that $a^2 + 2bc = 1$ and $b^2 + 2ca = 2012$, find all possible values
of $c^2 + 2ab$.

Proposed by Titu Andreescu.

Solution:

Subtracting the two given equations, we have $(b-a)(b+a-2c)=2011$, so we get $$\left\{ \begin{array}{lll} b & = & a \pm 1 \\ 2c & = & b+a \mp 2011 \end{array} \right. \qquad \left\{ \begin{array}{lll} b & = & a \pm 2011 \\ 2c & = & b+a \mp 1. \end{array} \right.$$
Substituting these values into the equation $a^2+2bc=1$, we obtain the four equations

(i) $3a^2-2008a-2011=0$
(ii) $3a^2+2008a-2011=0$
(iii) $3a^2+6032a+2010\cdot2011-1=0$
(iv) $3a^2-6032a+2010\cdot2011-1=0$.

Equation (i) gives $(a+1)(3a-2011)=0$, so $a=-1, b=0, c=-1006$.
Equation (ii) gives $(a-1)(3a+2011)=0$, so $a=1, b=0, c=1006$.
Equations (iii) and (iv) have no solution since their discriminant is negative.
In conclusion, $c^2+2ab=1006^2=1012036$.