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Let $x,y$ be real numbers such that :

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

Prove that :

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

I tried taking $x=y$. It simplifies everything a lot. But I'm not able to progress when both $x$ and $y$ are in the same equation.

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    Have you tried expanding? That is, have you tried multiplying it out? $(A+B)(C+D)=AC+AD+BC+BD$ – 2017-01-12
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    That does not seem to help a lot ... at least for me. – 2017-01-12
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    Rationalizing might help. – 2017-01-12
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    @guest, how would you rationalize this? I'm familiar with rationalizing e.g. complex numbers but not sure what you mean in this context. – 2017-01-12
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    [Using Approach0](https://approach0.xyz/search/?q=%24(x%2B%20%7B%5Csqrt%7B1%2Bx%5E2%7D%7D%20)%20(%20y%2B%20%7B%5Csqrt%20%7B1%2By%5E2%7D%7D)%20%3D%201%24&p=1) you can find some questoins which might be related: [Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$](http://math.stackexchange.com/q/1118742) and [$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find $(x+y)$.](http://math.stackexchange.com/q/1794502). – 2017-01-12
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    Thanks a lot @MartinSleziak – 2017-01-12

1 Answers 1

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Let $x =\sinh(a), y=\sinh(b)$.

Then $$(\sinh(a)+\cosh(b)) (\cosh(a) +\sinh(b))=1 \\ \sinh(a)\cosh(a)+\sinh(b)\cosh(b)+\cosh(a)\cosh(b)+\sinh(a)\sinh(b)=1 \\ \sinh(a+b)+\cosh(a+b)=1 $$

Now, $$1=\cosh^2(a+b)-\sinh^2(a+b)=\left( \sinh(a+b)+\cosh(a+b)\right)\left( \cosh(a+b)-\sinh(a+b)\right) \\=\left( \cosh(a+b)-\sinh(a+b)\right)$$ Therefore $$\sinh(a+b)+\cosh(a+b)=1\\ \cosh(a+b)-\sinh(a+b)=1$$ and hence $$\cosh(a+b)=1 \\ \sinh(a+b)=0$$

which proves that $a+b=0$.

Therefore $y=-x$.

From here it is trivial.

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    I don't know much about hyperbolic functions, but there is one thing for sure : your result "$y= - x$" proves that $y^2 = x^2$, which helps in substituting $y^2$ for $x^2$ in the first equation to yield the second equation. – 2017-01-12
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    Briliant idea!! – 2017-01-12
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    @QUANTUM You can alternately try the parametrisation $x=\tan(a), y=\tan(b)$ and try to reduce it. I used the $\sinh(a), \cosh(a)$ as the identities involving hyperbolic sin and cos are usually known better than the ones involving secant and tangent. – 2017-01-12