Square root and Square - Brothers!

Nice try, but please prove this problem!

We have:

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

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

So:

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

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

Calculate $2$ sides of $2$ equations $(*)$ and $(**)$ , erase all common terms, we have:

$x+y$ = $-x-y$

$\Leftrightarrow 2x+2y=0$

$\Leftrightarrow x+y=0$

Hence, $x+y=0$

Supposing $x$ and $y$ reals, we can make the following substitution: $x=\tan{\alpha}, y=\tan{\beta}; \alpha, \beta \in \left[0,\frac{\pi}{2}\right)$ $\left(\tan{\alpha}+\sqrt{\tan{\alpha}^{2}+1}\right)\left(\tan{\beta}+\sqrt{\tan{\beta}^{2}+1}\right)=1$ $(\tan{\alpha}+\sec{\alpha})(\tan{\beta}+\sec{\beta})=1$ $\tan{\alpha}\tan{\beta}+\tan{\alpha}\sec{\beta}+\sec{\alpha}\sec{\beta}+\sec{\alpha}\sec{\beta}=1$ $\sin{\alpha}\sin{\beta}+\sin{\alpha}+\sin{\beta}+1=cos{\alpha}\cos{\beta}$ $\cos(\alpha + \beta)=2\sin{\frac{\alpha + \beta}{2}}\cos{\frac{\alpha - \beta}{2}}+1$ $\sin^{2}{\frac{\alpha + \beta}{2}}=-\sin{\frac{\alpha + \beta}{2}}\cos{\frac{\alpha - \beta}{2}}$ i) $\sin\left({- \frac{\alpha+\beta}{2}}\right)=\sin\left({- \frac{\pi - \alpha + \beta}{2}}\right)$ $\Rightarrow \alpha=\frac{-\pi}{2}+2k\pi$ or $\beta=\frac{-\pi}{2}+2k\pi$ (Out of the domain!) $$ ii) $\sin{\frac{\alpha+\beta}{2}}=0 \Rightarrow \alpha+\beta=2k\pi$ Therefore: $x+y=\tan\alpha+\tan\beta=\tan\alpha+\tan(2k\pi-\alpha)=tan\alpha - tan\alpha =\boxed{0}.$

Multiply the equation out to get $xy+x\sqrt{ y^2+1}+y \sqrt{x^2+1}+ \sqrt{(y^2+1)(x^2+1)}=1$

Square both sides to get $x^2y^2+x^2(y^2+1)+y^2(x^2+1)+(x^2+1)(y^2+1)=1$

Simplify to get $4x^2y^2+2x^2+2y^2+1=1$

Subtract 1 from both sides to get $4x^2y^2+2x^2+2y^2 = 0$

The only way for this equation to be true is if $x^2=y^2=0$ and only $0^2=0$

So since $x=y=0$ the solution is $0+0=**0**$

There are only 2 possible solution of (x,y) namely x=y or x≠y we assume the solution is x=y,so (x+√(x^2 +1))(x+√(x^2+1))=1 (x+√(x^2 +1))^2=1 √(x^2 +1)=1-x Then we get the equation is x^2+1=x^2+1-2x So we can getting x=0 So the solution is x=y=0

X+Y=0

x=0; y=0; just check it out!