solved here and at more links from there

AlgebraFC wrote: We have\[a+\sqrt{1+a^2}=\frac{1}{b+\sqrt{1+b^2}}=\sqrt{1+b^2}-b\implies a+b=\sqrt{1+b^2}-\sqrt{1+a^2}\]and similarly\[b+\sqrt{1+b^2}=\frac{1}{a+\sqrt{1+a^2}}=\sqrt{1+a^2}-a\implies a+b=\sqrt{1+a^2}-\sqrt{1+b^2}.\]Adding the two equations gives $a+b=0$.

https://artofproblemsolving.com/community/c6h1682189p10727236 https://artofproblemsolving.com/community/c6h1202039p5913920 https://artofproblemsolving.com/community/c6h1202039p5913920 https://artofproblemsolving.com/community/c6h1713720p11062543 https://mp.weixin.qq.com/s/BwsKIyWArfGmTpQ21F2g_A https://math.stackexchange.com/questions/2574195/x-sqrtx21y-sqrty21-1-is-true-if-and-only-if-xy-0?noredirect=1