If $(x+\sqrt{x^2 + 1})(y+\sqrt{y^2 + 1})=p$, find $x+y$

Question

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

Answer 1

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.

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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 $x=y$, $$x+y=2x=\frac{p-1}{\sqrt{p}}$$

Answer 2

As other answers show, the problem is undetermined. However, if we introduce a constraint we can produce the particular solution the OP mentions in his comment.

By inspection the equation is symmetric in $x$ and $y,$ so a natural constraint to try is $x=y.$

If we impose this restriction, then $$x + \sqrt{x^2+1}=\sqrt{p}$$ $$\sqrt{x^2+1} = \sqrt{p}-x$$ $$x^2+1=p-2\sqrt{p}x+x^2$$ $$x=y={{p-1} \over {2\sqrt{p}}}$$ Then $$x+y={{p-1} \over \sqrt{p}}$$

Answer 3

Let us divide both sides of the equation by $(x + \sqrt{x^2 + 1})$. This gives $$y + \sqrt{y^2 + 1} = q,$$where we have defined $$q = \frac{p}{x + \sqrt{x^2 + 1}}.$$ Solving the equation for $y$ yields: $$y = \frac{q^2 - 1}{2q}.$$ Now all we have to do is add $x$ to this result for $y$, and you end up with the required expression. What you get is a function of $x$ and $p$.

[As others have shown, the result is not constant for a given $p$].

Answer 4

Let $x=\sinh a$ and $y=\sinh b$. Then we have $e^{a+b}=p\iff a+b=\ln p$. Since we're lacking a second equation, the system is undetermined. Depending on how we split $\ln p$ into two parts a and b, $\sinh a+\sinh b$ will have completely different solutions.