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

Question

For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$

Prove that $x+y=0.$

Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.

I see people are useing heavy artillery to prove such elementary problem. — nonuser, Sep 17 '17 at 11:56
I mean this is a problem for 9th grade and they definitive do not know what is $\mbox{arcsinh }$ or derivate is! — nonuser, Sep 17 '17 at 12:07

score 5 · Accepted Answer · answered Sep 17 '17 at 11:32

Multiplying by $(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$ we get

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

and thus $$(x+\sqrt{x^2 +1})(y+\sqrt{y^2 +1})=(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$$

hence

$$x(\sqrt{x^2 +1} +\sqrt{y^2 +1} )=-y(\sqrt{x^2 +1} +\sqrt{y^2 +1} )$$ therefore $$x=-y$$

score 0 · Answer 2 · answered Sep 17 '17 at 11:30

Since

$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1 / \cdot (x-\sqrt{x^2+1})$$

we get

$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x$$ so

$$\sqrt{x^2+1}-\sqrt{y^2+1}=x+y\;/^2$$

and thus $$-\sqrt{y^2+1}\cdot \sqrt{x^2+1}+1= xy$$

so $$x^2y^2 +x^2+y^2+1=x^2y^2-2xy+1$$

and finaly $(x+y)^2=0$ so $x+y=0$

score 0 · Answer 3 · answered Sep 17 '17 at 11:30

We have that $\mbox{arcsinh } x = \ln(x+\sqrt{x^2+1})$. If we take logs of both sides of the equation we have

$$\ln(x+\sqrt{x^2+1}) + \ln(y+\sqrt{y^2+1}) = 0$$

$$\mbox{arcsinh }x = -\mbox{arcsinh } y.$$

Then since $\mbox{arcsinh }$ is an odd function, we have $x=-y$.

score 0 · Answer 4 · answered Sep 17 '17 at 11:42

Take natural logarithm to the both sides, and it becomes

$\ln(x+\sqrt{x^2+1})+ln(y+\sqrt{y^2+1})=0$

Which can be written as

$\sinh^{-1}x+\sinh^{-1}y=0$

Since $\sinh^{-1}$ is odd and strictly increasing, there is

$\sinh^{-1}x=-\sinh^{-1}y=\sinh^{-1}-y$

Take $\sinh$ on both sides and Voila!

P.S. I didnt learn maths in English, so please tell me if I used a wrong word

score -2 · Answer 5 · answered Sep 17 '17 at 12:01

Let $x = \tan\alpha$ and $y = \tan\beta$ for some $\alpha,\beta\in(-\pi/2,\pi/2)$. Then $\sqrt{x^2+1} = \sec\alpha$ and $\sqrt{y^2+1} = \sec\beta$, so the condition can be rewritten as $$(\tan\alpha+\sec\alpha)(\tan\beta+\sec\beta) = 1. $$ Multiplying by $\cos\alpha\cos\beta$ yields $$(\sin\alpha+1)(\sin\beta+1) = \cos\alpha\cos\beta. $$ Multiplying by $(1-\sin\alpha)(1-\sin\beta)$ yields \begin{align} (1-\sin^2\alpha)(1-\sin^2\beta) &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\ \implies\cos^2\alpha\cos^2\beta &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\ \implies \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta). \end{align} Thus \begin{align} (1+\sin\alpha)(1+\sin\beta) = \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta) \\ \implies 1+\sin\alpha+\sin\beta+\sin\alpha\sin\beta &= 1-\sin\alpha-\sin\beta+\sin\alpha\sin\beta \\ \implies 2(\sin\alpha+\sin\beta) &= 0. \end{align} This implies that $\alpha = -\beta$, and hence $x = \tan\alpha = \tan(-\beta) = -\tan\beta = -y$, as desired.

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

5 Answers5