Mathematical Reflections 2012, Issue 4 - Problem O235

Problem:
Solve in integers the equation $$xy-7\sqrt{x^2+y^2}=1.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $xy \geq 1$. By symmetry, we reduce to $x,y > 0$. Rewriting the equation in the form $(xy-1)^2=49(x^2+y^2)$, we put $xy=7t+1, t \geq 0$, so that $x^2+y^2=t^2$. Then $$(x+y)^2=x^2+y^2+2xy=t^2+2(7t+1)=(t+7)^2-47,$$ which gives $$(t+7-x-y)(t+7+x+y)=47.$$
So, we must solve the systems
$$\left\{\begin{array}{rcl} t+7-x-y & = & 1 \\ t+7+x+y & = & 47 \end{array} \right. \qquad \left\{\begin{array}{rcl} t+7-x-y & = & -47 \\ t+7+x+y & = & -1 \end{array} \right.$$
Solving the first, we get $t=17$ and $x+y=23$, solving the second we get $t=-31<0$, i.e. no solution. So we have $x+y=23, xy=120$, which gives $x=15,y=8$ and $x=8, y=15$. In conclusion, the solutions are $$(15,8),(8,15),(-15,-8),(-8,-15).$$

Mathematical Reflections 2012, Issue 4 - Problem U237

Problem:
Let $\mathcal{H}$ be a hyperbola with foci $A$ and $B$ and center $O$. Let $P$ be an arbitrary point
on $\mathcal{H}$ and let the tangent of $\mathcal{H}$ through $P$ cut its asymptotes at $M$ and $N$. Prove that $PA + PB = OM + ON$.

Proposed by Luis Gonzalez.

Solution:

Let $\mathcal{H}: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, so that $O=(0,0)$. By symmetry, suppose that $P$ is on the first quadrant. Then $P=\left(k,\dfrac{b}{a}\sqrt{k^2-a^2}\right)$, where $k \geq a$ is an arbitrary real number. The equation of the tangent of $\mathcal{H}$ at $P$ is $\dfrac{kx}{a^2}-\dfrac{\dfrac{b}{a}\sqrt{k^2-a^2}y}{b^2}=1$, so intersecting this line with the asymptotes $y=\pm \dfrac{b}{a}x$ we get the points $$M=\left(k+\sqrt{k^2-a^2},\dfrac{b}{a}(k+\sqrt{k^2-a^2})\right), \qquad N=\left(k-\sqrt{k^2-a^2},-\dfrac{b}{a}(k-\sqrt{k^2-a^2})\right).$$ Then
$$OM+ON=\dfrac{2k}{a}\sqrt{a^2+b^2}=2ke,$$ where $e=\sqrt{a^2+b^2}/a$ is the eccentricity of the hyperbola. Since the foci have coordinates $A=(-c,0), B=(c,0)$ where $c=\sqrt{a^2+b^2}$, we have
$$\begin{array}{lcl}PA+PB & = &\sqrt{(k+c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}+\sqrt{(k-c)^2+\dfrac{b^2}{a^2}(k^2-a^2)} \\ & = & \dfrac{4kc}{\sqrt{(k+c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}-\sqrt{(k-c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}}\\ &=&\dfrac{2kc}{a}\\&=&2ke \end{array}$$
and the statement follows.

Mathematical Reflections 2012, Issue 4 - Problem U236

Problem:
Let $f(X)$ be an irreducible polynomial in $\mathbb{Z}[X]$. Prove that $f(XY)$ is irreducible in $\mathbb{Z}[X,Y]$.

Proposed by Mircea Becheanu.

Solution:

If $f(X)$ is a constant polynomial, the statement is trivial. So, assume that $\deg f(X) > 0$. Since $f(X)$ is irreducible, then must be primitive. Assume that $f(XY)$ is reducible in $\mathbb{Z}[X,Y]$. Then, there are two non constant polynomials $g(X,Y), h(X,Y) \in \mathbb{Z}[X,Y]=\mathbb{Z}[X][Y]$ such that $f(XY)=g(X,Y)h(X,Y)$. It cannot be $f(XY)=g(X)h(Y)$, otherwise $f(0)=g(0)h(Y)$, so $h(Y)$ must be constant. Then both $g(X,Y)$ and $h(X,Y)$ must contain both $X$ and $Y$. But $$f(X)=g(X,1)h(X,1)=\overline{g}(X)\overline{h}(X), \qquad \overline{g}(X), \overline{h}(X) \in \mathbb{Z}[X]$$ is the product of two non constant polynomials, i.e. $f(X)$ is reducible, contradiction.

Mathematical Reflections 2012, Issue 4 - Problem S235

Problem:
Solve the equation $$\dfrac{8}{\{x\}}=\dfrac{9}{x}+\dfrac{10}{[x]},$$ where $[x]$ and $\{x\}$ denote the greatest integer less or equal than $x$ and the fractional part of $x$, respectively.

Proposed by Titu Andreescu.

Solution:

Clearly $x \notin \mathbb{Z}, x \not \in (0,1)$ and from $\dfrac{8}{\{x\}} > 0$, we gather $x > 1$. Since $0 < \{x\} < 1$ and $0 < [x] \leq x$, we have $$8 < \dfrac{8}{\{x\}} \leq \dfrac{19}{[x]},$$ so $[x]=1,2$. Clearing the denominators and using the fact that $\{x\}=x-[x]$ we obtain the equation $$10x^2-9x[x]-9[x]^2=0.$$
(i) If $[x]=1$, we get $10x^2-9x-9=0$, which gives $x=\dfrac{3}{2},-\dfrac{3}{5}$, i.e. $x=\dfrac{3}{2}$. 

(ii) If $[x]=2$, we get $10x^2-18x-36=0$, which gives $x=3,-\dfrac{6}{5}$, i.e. no solution.

So, $x=\dfrac{3}{2}$ is the only solution to the given equation.

Mathematical Reflections 2012, Issue 4 - Problem J239

Problem:
Let $a$ and $b$ be real numbers so that $2a^2 + 3ab + 2b^2 \leq 7$. Prove that $$\max\{2a+b,2b+a\} \leq 4.$$

Proposed by Titu Andreescu.

Solution:

Suppose that $a \geq b$, so that $\max\{2a+b,2b+a\}=2a+b$. We argue by contradiction. Suppose that for some real numbers $a,b$ we have $2a+b>4$, i.e. $a>\dfrac{4-b}{2}$. Then
$$7 \geq 2a^2 + 3ab + 2b^2 > b^2+2b+8,$$ which gives $(b+1)^2 < 0$, contradiction.

Mathematical Reflections 2012, Issue 4 - Problem J237

Problem:
Prove that the diameter of the incircle of a triangle $ABC$ is equal to $\dfrac{AB-BC+CA}{\sqrt{3}}$ if and only if $\angle BAC = 60^{\circ}$.

Proposed by Titu Andreescu.

Solution:
Let $r$ be the radius of the incircle of $\triangle ABC$ and let $\alpha=\angle BAC$. Clearly $0^{\circ} < \alpha < 180^{\circ}$ and $$2r=\dfrac{2AB\cdot CA \sin \alpha}{AB+BC+CA}.$$ Then $$2r=\dfrac{AB-BC+CA}{\sqrt{3}} \iff AB^2+CA^2+2AB\cdot CA - BC^2=2\sqrt{3}AB \cdot CA \sin \alpha.$$ Since $BC^2=AB^2+CA^2-2AB\cdot CA \cos \alpha$, we obtain $$2r=\dfrac{AB-BC+CA}{\sqrt{3}} \iff 1 + \cos \alpha = \sqrt{3} \sin \alpha,$$ i.e. if and only if $\sin (\alpha-30^{\circ})=\dfrac{1}{2},$ which gives $\alpha=60^{\circ}$.

Mathematical Reflections 2012, Issue 4 - Problem J235

Problem:
In the equality $\sqrt{ABCDEF}=DEF$, different letters represent different digits. Find the six-digit number $ABCDEF$.

Proposed by Titu Andreescu.

Solution:
From the given equality, it must be $ABCDEF^2-DEF \equiv 0 \pmod {1000}$, and since $ABCDEF \equiv DEF \pmod{1000}$, we have $DEF^2-DEF=DEF(DEF-1) \equiv 0 \pmod{1000}$. Since $DEF$ and $DEF-1$ are coprime and $1000=2^3\cdot5^3$, one between this two numbers must be odd and divisibile by $5^3$ and the other must be divisible by $2^3$. Since $DEF$ and $DEF-1$ are three digit numbers, $DEF \in \{125,375,625,875\}$ or $DEF-1 \in \{125,375,625,875\}$. In the first case, $DEF-1$ is divisible by $8$ if and only if $DEF=625$, in the second case $DEF$ is divisible by $8$ if and only if $DEF-1=375$. So, $DEF=625,376$, but
$$625^2=390625 \qquad 376^2=141376,$$ i.e. the only number which satisfies the required conditions is $625$.

KöMaL (Metresis) 1894, Problem 8

Problem:
If $2 \cos \vartheta = u+\dfrac{1}{u}$, prove that $2 \cos n \vartheta = u^n+\dfrac{1}{u^n}$ for every positive integer $n$.

Solution 1:
We prove the statement by induction on $n \geq 2$. If $n=2$ we have $$2 \cos 2\vartheta =4 \cos^2 \vartheta - 2=\left(u+\dfrac{1}{u} \right)^2 - 2=u^2+\dfrac{1}{u^2}.$$ Assume that the statement is true for every $2 \leq k \leq n$. Then, $$\begin{eqnarray*} 2 \cos (n+1)\vartheta &= &2[2\cos n \vartheta \cos \vartheta - \cos (n-1)\vartheta]\\ &= & 2\left[\left(u^n+\dfrac{1}{u^n}\right)\dfrac{1}{2}\left(u+\dfrac{1}{u}\right)-\dfrac{1}{2}\left(u^{n-1}+\dfrac{1}{u^{n-1}}\right)\right] \\ & = & u^{n+1}+\dfrac{1}{u^{n+1}}, \end{eqnarray*}$$ and the statement follows.

Solution 2:
From the given equality, we obtain $u^2-(2\cos \vartheta) u + 1=0$, and solving this equation in $u$ we get $u=\cos \vartheta \pm i \sin \vartheta=e^{\pm i \vartheta}$. Then, for every positive integer $n$
$$2 \cos n \vartheta = e^{\pm in\vartheta}+e^{\mp in \vartheta}=u^n+u^{-n}=u^n+\dfrac{1}{u^n}$$ as we wanted to prove.

KöMaL (Metresis) 1894, Problem 2

Problem:
Find all positive integers $N$ divisible only by $2$ and by $3$ such that $N^2$ has three times the number of divisors of $N$.

Solution:
Let $N=2^a 3^b$, where $a,b \in \mathbb{N}^*$. Then $N^2 = 2^{2a}3^{2b}$. So, it must be $$3(a+1)(b+1)=(2a+1)(2b+1) \iff ab-a-b-2=0 \iff (a-1)(b-1)=3.$$ Since $a-1 \geq 0$ and $b-1 \geq 0$, we obtain $$\left\{ \begin{array}{rcl} a-1 & = & 1 \\ b-1 & = & 3 \end{array} \right. \qquad \textrm{or} \qquad \left\{ \begin{array}{rcl} a-1 & = & 3 \\ b-1 & = & 1 \end{array} \right.$$ i.e. $a=2, b=4$ or $a=4, b=2$. Then all the positive integers required are $N_1=2^2 3^4=324$ and $N_2=2^4 3^2=144$.

KöMaL (Metresis) 1894, Problem 1

Problem: 
Find all the two digit numbers $\overline{ab}$ such that $\overline{ababab}+1$ is a perfect cube.

Solution:
Let $x=\overline{ab}$ be the two digit number. Then, $101010x+1=n^3$, for some positive integer $n$. Clearly, $10^6 \leq n^3 < 10^7$, i.e. $100 \leq n \leq 215$. Moreover,
$$101010x=n^3-1.$$ Since $101010=30\cdot3367$, it's easy to see that $n^3-1 \equiv 0 \pmod{30}$ if and only if $n \equiv 1 \pmod{30}$, i.e. $n-1$ is divisible by $30$. So $n \in \{121,151,181,211\}$ and an easy check shows that only $n=211$ works. In this case, $x=93$.