Given the equation below: $$\sqrt{a+\sqrt b}=\sqrt x+\sqrt y$$ Show that: $$\sqrt{a-\sqrt b}=\sqrt x-\sqrt y$$
I saw the above problem. However, I can't figure out how to manipulate the above. Furthermore, are these equations even true? If $a=21$, $b=16$, $x=9$ and $y=4$ then the other equation does not hold true. Is the question wrong?
The given equation is:
$$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$$
Here we have to consider two cases. $\mathbf{CASE-I}$ If $b$ is not a perfect square then after squaring both sides ${(\sqrt{a+\sqrt{b}})}^2={(\sqrt{x}+\sqrt{y})}^2 \Rightarrow a+\sqrt{b}=x+y+2 \sqrt{xy} \Rightarrow a-x-y=2 \sqrt{xy}-\sqrt{b}$
As $a-x-y$ is an integer so $2 \sqrt{xy}-\sqrt{b}$ should also be an integer. But they are not perfect squares so they are same thus $2 \sqrt{xy}=\sqrt{b}$ Hence $a-\sqrt{b}=x+y-2 \sqrt{xy}$ So $\sqrt{a-\sqrt{b}}=\sqrt{x}-\sqrt{y}$
$\mathbf{CASE-II}$
If $b$ is a perfect square then after squaring both sides ${(\sqrt{a+\sqrt{b}})}^2={(\sqrt{x}+\sqrt{y})}^2 \Rightarrow a+\sqrt{b}=x+y+2 \sqrt{xy}$ As $b$ is a perfect square so as $xy$. Now, $a+\sqrt{b}-2 \sqrt{xy}=x+y-2 \sqrt{xy} \Rightarrow \sqrt{a+\sqrt{b}-2 \sqrt{xy}} =\sqrt{x+y-2 \sqrt{xy}} =\sqrt{x}-\sqrt{y}$ To get the form $\sqrt{a-\sqrt{b}}$ we need to have $b= xy$ which is not true in your example
As several of the other comments have stated, the claim is wrong in general. I believe the idea here is that by choosing appropriate $x$ and $y$, we may be able to simplify an expression of the form $\sqrt{p+q\sqrt{r}}$.
Note that squaring both sides of the given equation yields
$$\begin{align} a + \sqrt{b} &= x + y + 2\sqrt{xy} \\[2mm] &= x + y + \sqrt{4xy} \end{align}$$
Now if we can find naturals $x$ and $y$ such that $\; a = x + y \;$ and $\; b = 4xy \;$ then we would be able to simplify $\sqrt{a+\sqrt{b}}$.
In particular, to simplify the given $\sqrt{84+16\sqrt{17}}$ we first note that we can factor out a 2 and rewrite as $2\sqrt{21+4\sqrt{17}} = 2\sqrt{21+\sqrt{4 \cdot 68}}$; thus we are looking for natural solutions to the system $x+y=21$ and $xy=68$.
This is relatively straightforward:
$$\begin{align} (x-y)^2 &= (x+y)^2 - 4xy \\[2mm] &= 21^2 - 4(68) \\[2mm] &= 169 \\[2mm] x-y &= 13 \end{align}$$
which then combining with $x+y=21$ easily yields $x=17$ and $y=4$.
Thus
$$\begin{align} \sqrt{84+16\sqrt{17}} &= 2\sqrt{21+4\sqrt{17}} \\[2mm] &= 2 \left( \sqrt{17} + \sqrt{4} \right) \\[2mm] &= 4 + 2\sqrt{17} \end{align}$$
Addendum:
Initially I felt the claim was useless in answering the second part; however after I wrote the above solution I realized it could be used in the following way: if we let, as above, $a = x + y$ and $b = 4xy$, then
$$\begin{align} a - \sqrt{b} &= x + y - \sqrt{4xy} \\[2mm] &= x + y - 2\sqrt{xy} \\[2mm] &= \left( \sqrt{x} - \sqrt{y} \right)^2 \end{align}$$
and therefore
$$\begin{align} x - y &= \left( \sqrt{x} + \sqrt{y} \right) \left( \sqrt{x} - \sqrt{y} \right) \\[2mm] &= \left( \sqrt{a + \sqrt{b}} \right) \left( \sqrt{a - \sqrt{b}} \right) \\[2mm] &= \sqrt{a^2 - b} \end{align}$$
which would then yield the same $x-y=13$ as above.
If we make the assumption that $a, b, x, y$ are integers (or even rational), then see Wtsm's solution (with slight modifications). Otherwise, read on.
Lemma: The claim is true iff $ \sqrt{a^2 - b} = x-y$.
Proof: Given that condition, $ \sqrt{a - \sqrt{b}} = \frac{ \sqrt{ a^2 - b } } { \sqrt{ a - \sqrt{b} }} = \frac{ x-y}{\sqrt{x} + \sqrt{y} } = \sqrt{x} - \sqrt{y} $.
The converse follows by multiplying the 2 equations and using the same algebraic manipulation.
Corollary: For a given $a, b$, if $x, y$ are not restricted to integers, then we can find numerous solutions sets to $ \sqrt{ a + \sqrt{b} } = \sqrt{x} + \sqrt{y}$, but only one of them will also satisfy $ \sqrt{a^2 - b} = x-y$.
In particular, for your chosen values of $a=21$, $b=16$, $x=9$ and $y=4$, since $a^2 - b = 425 \neq 25 = (x-y)^2$, so the conclusion will not hold.
The claim is true by isomorphism: that is you can replace $\sqrt b$ by $-\sqrt b$ as long as $by$ or $bx$ is itself a square.
