If $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$, find $x+y$ I was given this factorization problem and I tried many things, but couldn't solve it. Can someone, please, give me a hint? 

If $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$, find $x+y$. Here $x, y$ and $p$ are real numbers.

Thank you

This is a bad question, since:

The identity $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$ does not determine $x+y$. 

Example: Let $p=4$, then $(x,y)=\left(0,\frac{15}8\right)$ and $(x,y)=\left(\frac34,\frac34\right)$ are both solutions, for $x+y=\frac{15}8$ and $x+y=\frac32$ respectively.

If one wishes to add the constraint that $$x=y$$ as the OP seems to be willing to do, then one should ask at the onset to solve $$x+\sqrt{x^2 + 1}=\sqrt{p}$$ Then $$(x+\sqrt{x^2+1})\cdot(x-\sqrt{x^2+1})=-1$$ hence every solution $x$ is such that $$2x=(x+\sqrt{x^2+1})+(x-\sqrt{x^2+1})=\sqrt{p}-\frac1{\sqrt{p}}$$ that is, since once again one assumed $x=y$, $$x+y=2x=\frac{p-1}{\sqrt{p}}$$
