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Tuesday, April 3, 2012

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.