Një pabarazi si çdo tjetër // An Inequality like any other

(Albanian)          

                     A photo of a solution from Facebook looks like this. She is trying to  solve a problem proposed by Bogdan ENESCU, from an unspecified source. The same man raised the problem a year ago.

        Statement:

                           Let $x,y \in \mathbb{R}.$ Prove that

1. If $(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1$, then $x+y=0;$

2. If $(x+\sqrt{1+y^2})(y+\sqrt{1+x^2})=1$, then $x+y=0.$

Many solutions, the most elegant using the hyperbolic trigonometric function substitution, appear in the comments. Here we only point out an algebraic inequality that solves point 2. of the problem.

  

          LEMMA The inequality 

$$\sqrt{1+x^2} \cdot \sqrt{1+y^2}\geqslant |1-x\cdot y| \tag{1}$$

           holds, for any $x,y \in \mathbb{R}$; the "$=$" sign occurs if and only if $x+y=0.$

          Proof of the lemma

$(x+y)^2 \geqslant 0\;\Leftrightarrow \;x^2+2xy+y^2 \geqslant 0\;\Leftrightarrow \;1+x^2+y^2+x^2 \cdot y^2 \geqslant 1-2xy+x^2\cdot y^2\;$

$\Leftrightarrow \;(1+x^2)(1+y^2)\geqslant (1-xy)^2\;\Leftrightarrow \;(1)$.

$\square$

        Now $(x+\sqrt{1+y^2})(y+\sqrt{1+x^2})=1\;\Rightarrow$

$$\Rightarrow\;xy+y\sqrt{1+y^2}+x\sqrt{1+x^2}+\sqrt{1+x^2}\sqrt{1+y^2}=1\;\Rightarrow$$

$$\Rightarrow\;y\sqrt{1+y^2}+x\sqrt{1+x^2}=1-xy-\sqrt{1+x^2}\sqrt{1+y^2}\;\overset{squaring}{\Rightarrow}$$

$$\Rightarrow\;y^2(1+y^2)+x^2(1+x^2)+2xy\sqrt{1+x^2}\sqrt{1+y^2}=$$

$$=1+x^2y^2+(1+x^2)(1+y^2)-2xy-2\sqrt{1+x^2}\sqrt{1+y^2}+2xy\sqrt{1+x^2}\sqrt{1+y^2}\;\Rightarrow$$

$$\Rightarrow\;x^4+y^4-2x^2y^2=2-2xy-2\sqrt{1+x^2}\sqrt{1+y^2}\;\Rightarrow$$

$$\Rightarrow\;1-xy-\sqrt{1+x^2}\sqrt{1+y^2}=\frac{(x^2-y^2)^2}{2}.\tag{2}$$

But, from the Lemma we have

$1-xy\leqslant |1-xy|\leqslant \sqrt{1+x^2}\sqrt{1+y^2}$. So $1-xy-\sqrt{1+x^2}\sqrt{1+y^2}\leqslant 0$

and then from (2) it results $\frac{(x^2-y^2)^2}{2}\leqslant 0$. Hence $x^2=y^2$.

If $x=y$, then 2. is written $(x+\sqrt{1+x^2})^2=1$ which is successively equivalent to $x^2+(1+x^2)+2x\sqrt{1+x^2}=1\;\Leftrightarrow\;x^2+x\sqrt{1+x^2}=0$, hence $x=0=y.$

If $x=-y$, then $(x+\sqrt{1+y^2})(x+\sqrt{1+x^2})=(x+\sqrt{1+x^2})(-x+\sqrt{1+x^2})=(1+x^2)-x^2=1.$ In both cases 2. is fulfilled.

         Tanks to Doru Stoica for this solution.

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Solving Pytagorean Equation $x^2+y^2=z^2$ / حل معادله فیثاغورث

              (Persian) به نشانه تحسین علیرضا صالحی گل سفیدی که کتابش بارها به من روحیه داده است.

                 As a sign of admiration for Alireza Salehi Golsefidy whose book has inspired me many times.

                A Pytagorean_triple consist of three positive integers $\;a,\;b\;$ and $\;c,\;$ such that

$$a^2+b^2=c^2\;. \tag{P}$$

          This is how the Wikipedia article about them begins. Being dissatisfied with the way the subject is presented in various works, I will present the version that seemed to me the most correct and complete.

          The presentation from page 29, from this source :

COHEN Henri

Explicit Metods for Solving Diophantine Equations

f.e., f.l., 2006

seemed to me the best.

          

          If the triplet $(a,b,c)$ verifies the equation $(P)$, then the triplets $(\pm a,\pm b,\pm c),\; (b,a,c),\;(d\cdot a,d\cdot b,d \cdot c)\;-\;d\in \mathbb{Z}\;-$ also verifies $(P)$. Our luck is that, if the $\mathbf{gcd}(a,b)=1$, then $\mathbf{gcd}(a,b,c)=1$, and converselly. Thus, the general solution in integer numbers of the equation 

$$x^2+y^2=z^2 \tag{1}$$

is obtained from all the primitive solutions - that is, those for which $\mathbf{gcd}(x,y,z)=1$ - by multiplying by an arbitrary integer factor.

          On page 29, loc. cit. we read:

          "Proposition 5.1  The general solution to the equation (1) is given by the

          two disjoint parametrisations

$$(x,y,z)=(2st,s^2-t^2,\pm(s^2+t^2)) \tag{2}$$

     and

$$(x,y,z)=(s^2-t^2,2st,\pm(s^2+t^2)) \tag{3}$$

          where $s$ and $t$ are two coprime integers of opposite parity."[ and $s>0$, n.CiP]

     Remark CiP The addition of the right parentheses, made by us, is necessary because the pairs of parameter $(s,t)$ and $(-s,-t)$ give the same result. If we impose in the relations (2) and (3) as $s>t>0$, then all the solutions in non-zero natural numbers are obtained.

               We show in the table below some positive solutions obtained by formula (2); also, on a red background, solutions that violate the condition of parameters $s$ and $t$ to be of different parities (thus obtaining non-primitive solutions).

             In the proof of the Proposition 5.1, made also in the line of the cited work, we need the following

               LEMMA  Let $y$ and $z$ be two odd integers. Then

$$(y,z)=\left ( \frac{y+z}{2},\frac{y-z}{2}\right );$$

                   consequently      $(y,z)=1\;\Leftrightarrow\; \left (\frac{y+z}{2},\frac{y-z}{2}\right )=1.$

           (Here, $(u,v)$ denotes the greatest common divisor of the numbers $u$ and $v$.)

                Proof of Lemma  Let $d:=(y,z)\;$; from Bézout's Lemma there exist integers $u$ and $v$ such that $y \cdot u+z \cdot v=d$. From here $$2d=2yu+2zv=(y+z)\cdot (u+v)+(y-z) \cdot (u-v),$$ so $d=\frac{y+z}{2}\cdot (u+v)+\frac{y-z}{2} \cdot (u-v)$, and again with Bézout's Lemma, we have

$$d_1:= \left (\frac{y+z}{2},\frac{y-z}{2} \right )\;\mid \;d. \tag{4}$$

     Now, from $d=(y,z)$ result $y=d \cdot u_1,\;z=d \cdot v_1$, with $(u_1,v_1)=1.$ Hence $\frac{y \pm z}{2}=d\cdot \frac{u_1 \pm v_1}{2}$, so, $u_1$ and $v_1$ being odd, $\;d\;\mid\;\frac{y \pm z}{2}\;$ so we must have

$$d\;\mid \; d_1. \tag{4'}.$$

Relations (4) and (4') show that $d=d_1$.

(end of proof Lemma)$\square$

                   Proof of Proposition 

                   We will look for solutions in positive integers. 

               The $x$ and $y$ cannot both be odd, because then $x^2+y^2$ has the form $4k+2$, which cannot be a perfect square. Then exactly one is odd and the other, say $x$, is even.

          We rewrite the equation (1) like this: $\left ( \frac{x}{2} \right )^2=\underset{q}{\underbrace{\frac{z-y}{2}}} \cdot \underset{p}{\underbrace{\frac{z+y}{2}}}.$ The $y$ and $z$ being 

coprime, according to the Lemma, $q=\frac{z-y}{2}$ and $p=\frac{z+y}{2}$ are also coprime. Then, in the last equation we must have both $\frac{z \pm y}{2}$ be squares. 
$$q=\frac{z-y}{2}=t^2\,,\,\;p=\frac{z+y}{2}=s^2 \tag{5} .$$

The simple observation that $y=p-q$ and $z=p+q$ shows that $p,\;q$ must be of different parities (otherwise $y$ and $z$ would be even); the same are $s,\;t$.

          From (5) and further from $(x/2)^2=p \cdot q$ results the formulas (2).

$\blacksquare$

A Problem for IMO, which does not deserve to be considered / Проблем за ИОМ, који не заслужује да буде разматран

                    In the CRUX mathematics magazine, [2000 : 24] the review of a mathematics book, related to the International Olympiads, appeared.

In addition to the given problems, there is a list of problems that are discussed in the competition jury, but which no longer see the light of day.

          Today we will deal with Problem #2 from the list in the previous image.

"Suppose $\tan x =\frac{p}{q}$, where $p$ and $q$ are positive integers and $q \neq 0.$

Prove that the number $\tan y$ for wich $\tan 2y=\tan 3x$ is rational only

      when $p^2+q^2$ is a perfect square."                                                                                                                        

          The solution is too simple for the problem to be at the level of the problems given at the IOM (even from the early 60's).

Solution CiP

                    Let $\tan x =\frac{p}{q}.$ With the triple angle formula $\tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2x}$ we obtain $\tan 3x=\frac{p(3q^2-p^2)}{q(q^2-3p^2)}.$

     We also apply the double angle formula $\tan 2y=\frac{2\tan y}{1-\tan^2y}.$

          The  equation $\tan 2y=\tan 3x$ is written

$$\frac{2\tan y}{1-\tan^2y}=\frac{p(3q^2-p^2)}{q(q^2-3p^2)}.$$

Or equivalent

$$p(3q^2-p^2)\cdot \tan^2y+2q(q^2-3p^2) \cdot \tan y-p(3q^2-p^2)=0.$$

In order for the previous equation, with the unknown $\tan y$, to admit rational number solutions, we must have $\Delta$ to be a perfect square. Here $\Delta=b^2-4ac$ is the discriminant of the quadratic equation $a\cdot t^2+b\cdot t+c=0.$ Since $b=2\cdot b'$ is an even number, it is enough to consider the discriminant "halved" $\Delta '=(b')^2-ac$. For us
$$\Delta ' =q^2(q^2-3p^2)^2+p^2(3q^2-p^2)^2=$$

$$=q^6-6p^2q^4+9p^4q^2+9p^2q^4-6p^4q^2+p^6=$$

$$=p^6+3p^4q^2+3p^2q^4+q^6=(p^2+q^2)^3=(p^2+q^2)^2 \cdot (p^2+q^2).$$

          Now it is obvious that $\Delta '$ is a perfect square if and only if $p^2+q^2$ is a perfect square.

$\blacksquare$

     

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ЛЕПА ЈЕДНАЧИНА СА ФУНКЦИЈОМ ПУНИХ ДЕЛОВА

(Sârbă) 

A BEAUTIFUL EQUATION WITH FLOOR FUNCTION

 

               $[\alpha]$ denotes the largest integer less than or equal to $\alpha$ (floor function). An equation that uses all the numbers from 1 to 9, considered by us "beautiful" is

$$\left [  \frac{2x+1}{3}\right ]+\left [ \frac{4x+5}{6} \right ]=\frac{7x-8}{9}.$$

ANSWER CiP

$$x \in \left \{ -\frac{19}{7},\;-\frac{10}{7}\right \}$$

 Solution CiP

               We use characterization $[\alpha] \leqslant \alpha <[\alpha]+1$, so
$$[\alpha]=k\; \Leftrightarrow \;k \in \mathbb{Z}\;and\; k\leqslant \alpha<k+1.$$

          The left side of the equation is an integer, so we must have $\frac{7x-8}{9}=k,\;k\in\mathbb{Z}.$ Hence
$$x=\frac{9k+8}{7}.\tag{1}$$

          Applying the equivalent to the definition inequalities  $\alpha-1<k\leqslant \alpha$ , we have

$$\frac{2x+1}{3}-1<\left [ \frac{2x+1}{3} \right ] \leqslant \frac{2x+1}{3},\tag{2}$$

$$\frac{4x+5}{6}-1<\left [ \frac{4x+5}{6} \right ] \leqslant \frac{4x+5}{6}.\tag{3}$$

Adding these two relationships and considering the equation, we get

$$\frac{2x+1}{3}+\frac{4x+5}{6}-2<\frac{7x-8}{9}\leqslant \frac{2x+1}{3}+\frac{4x+5}{6}.$$

      The first inequality is successively equivalent to

$$\frac{2x+1}{3}+\frac{4x+5}{6}-\frac{7x-8}{9}<2\;\Leftrightarrow$$

$$\Leftrightarrow\;\frac{10x+37}{18}<2\;\Leftrightarrow\;x<-\frac{1}{10}.$$

The second inequality is successively equivalent to

$$\frac{2x+1}{3}+\frac{4x+5}{6}-\frac{7x-8}{9}\geqslant 0\;\Leftrightarrow$$

$$\Leftrightarrow\;\frac{10x+37}{18}\geqslant 0\;\Leftrightarrow \;x\geqslant -\frac{37}{10}.$$

     So we have  $-\frac{37}{10}\leqslant x <-\frac{1}{10}$ and considering the form (1) of $x$:

$$-\frac{37}{10}\leqslant \frac{9k+8}{7}<-\frac{1}{10}.$$

Solving these inequalities with the unknown $k$, we find $-\frac{113}{30}\leqslant k<-\frac{29}{30}.$ But the number $k$ being integer, we can only have $k \in \{-3,-2,-1\}$, obtaining for $x$, with the formula (1), respectively

$$x \in \left \{-\frac{19}{7}, -\frac{10}{7},-\frac{1}{7} \right \}.$$

      But the value $x=-\frac{1}{7}$ does NOT check the equation (it appeared as a "foreign solution", summing two relations (2) and (3), but which are not independent of each other). We get the answer.

$\blacksquare$

 

          REMARK CiP

                $1^{R}$. Noting that for $\alpha =\frac{2x+1}{3}$ we have $\alpha +\frac{1}{2}=\frac{4x+5}{6}$, the left side of the given equation has the form $[\alpha ]+\left [\alpha +\frac{1}{2} \right ]$. But we have an identity of Hermite, we can say as beautiful

$$[\alpha]+\left [ \alpha +\frac{1}{2} \right ]=[2\cdot \alpha].$$

Then our equation is written equivalently

$$\left [ 2\cdot \frac{2x+1}{3}\right ]=\frac{7x-8}{9}\;\Leftrightarrow\;\left [ \frac{4x+2}{3} \right ]=\frac{7x-8}{9}$$

which is a more common type in problem collections.

          Like before $x=\frac{9k+8}{7}$ and now it is enough to set the condition $k\leqslant \frac{4x+2}{3}<k+1\;\Leftrightarrow\;3k-2\leqslant 4x<3k+1\;\Leftrightarrow\;3k-2\leqslant \frac{36k+32}{7}<3k+1\;\Leftrightarrow$

$\Leftrightarrow\;-\frac{46}{15}\leqslant k<-\frac{5}{3}$, hence $k=-3$ or $k=-2$. We get the answer. The foreign solution stopped appearing.

(End Rem 1.)

                

               $2^{R}$. Our problem is suggested by Problem 2088 [1995: 307] from the journal CRUX_MATHEMATICORUM. (The notation in right brackets means the year 1995, page 307. More specifically, it is volume 21, number 9, from November 1995, under the heading "PROBLEMS"). I modified, for aesthetic reasons, only the term on the right side of the equation. The solution appeared under the heading SOLUTIONS, in their coding [1996: 334]. 

(End Rem 2.)