Find all $a$ and $b$ that make this value rational: $ \left(\sqrt 2 + \sqrt a\right)/\left(\sqrt 3 + \sqrt b\right)$ 
$$ \frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b}  $$ Find all integers $a$ and $b$ that make this value rational.

Looking at it, you will easily see that $a=3$, $b=2$ will make this value $1$. 
Are there more possible $a$ and $b$?
 A: Note that $b=3$ does not give any solution since $$\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt 3}=\frac{\sqrt{6}+\sqrt{3a}}{6}$$and $\sqrt{6}+\sqrt{3a}$ cannot be rational number (See this question for why). Therefore, we can rationalize it.
Rationalize the value and expanding gives$$\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b}=\frac{(\sqrt 2 + \sqrt a)(\sqrt 3 - \sqrt b)}{3-b}=\frac{\sqrt 6 - \sqrt{2b}+\sqrt{3a}-\sqrt{ab}}{3-b}$$which is rational if and only if $\sqrt 6 - \sqrt{2b}+\sqrt{3a}-\sqrt{ab}$ is rational. Let $\sqrt 6 - \sqrt{2b}+\sqrt{3a}-\sqrt{ab}=P$. I'll omit rational numbers in the following calculations because they don't change $P$'s rationality.
$$\sqrt 6+\sqrt{3a}=P+\sqrt{ab}+\sqrt{2b}$$
Squaring both sides,
$$2\sqrt{18a}=P'+2P\sqrt{ab}+2P\sqrt{2b}+2b\sqrt{2a}$$
Where $P'$ is remaining rational part of the equation. It is just $(6-2b)\sqrt{2a}-2P\sqrt{2b}=P'+2P\sqrt{ab}$, and squaring both sides again,
$$-8P(6-2b)\sqrt{ab}=P''+4PP'\sqrt{ab}$$
and therefore, $\sqrt{ab}$ is rational or, as an exceptional case, $-8P(6-2b)=4PP'$ or $PP'=4P(b-3)$.
Let's consider the case where $PP'=4P(b-3)$ first.

From $b \neq 3$, $P=(\sqrt 2 + \sqrt a)(\sqrt 3 - \sqrt b)\neq 0$. Therefore, $P'=4(b-3)$.
Squaring both sides of $\sqrt 6+\sqrt{3a}=P+\sqrt{ab}+\sqrt{2b}$ and considering rational term gives $P'=P^2+ab+2b-6-3a$. Applying $P'=4(b-3)$ and processing gives $P^2=(a-2)(3-b)$. That is, either $P=0,a=2$ or $a>2, b<3$ or $a<2, b>3$.
$P=0,a=2$ gives $b=3$, which we ruled out.
For $a>2, b<3$, either $b=0$, $b=1$ or $b=2$. $b=0$ means $\sqrt6+\sqrt{3a}=P$, so no solution.
$b=1$ means $P=\sqrt{2(a-2)}$, so $(\sqrt 2 + \sqrt a)(\sqrt 3 - 1)=\sqrt{2(a-2)}$. This case gives the solution $a=6$, but $\frac{\sqrt 2 + \sqrt 6}{\sqrt 3 + \sqrt 1}$ is not rational.
$b=2$ means $P=\sqrt{a-2}$, so $(\sqrt 2 + \sqrt a)(\sqrt 3 - \sqrt{2})=\sqrt{a-2}$. This case gives the solution $a=3$.
For $a<2, b>3$, either $a=0$ or $a=1$. $a=0$ means $\sqrt6-\sqrt{2b}=P$, which means $b=3$, which we ruled out.
$a=1$ means $P=\sqrt{b-3}$, so $(\sqrt 2 + 1)(\sqrt 3 - \sqrt{b})=\sqrt{b-3}$. This case gives the solution $b=3$, which we ruled out.
Therefore, one solution $(a, b)=(3, 2)$ exists in this case.

Now, what is left is the case where $\sqrt{ab}$ is rational.
From $(6-2b)\sqrt{2a}-2P\sqrt{2b}=P'+2P\sqrt{ab}$, $(6-2b)\sqrt{2a}-2P\sqrt{2b}$ is rational. Once again, by extension of above question, either $(6-2b)\sqrt{2a}$ and $2P\sqrt{2b}$ are rational or $(6-2b)\sqrt{2a}=2P\sqrt{2b}$ or $a(3-b)^2=P^2b$.
Firstly, let's consider the case $a(3-b)^2=P^2b$.
Since $b=0$ does not give any solution, we can divide both sides by $b$ to get $P=(3-b)\sqrt{\frac{a}{b}}$. Now we get$$(\sqrt2+\sqrt{a})(\sqrt3-\sqrt b)=(3-b)\sqrt{\frac{a}{b}}$$and since $b \neq 3$, we can divide each side by $\sqrt3-\sqrt b$ $$\sqrt2+\sqrt{a}=(\sqrt3+\sqrt b)\sqrt{\frac{a}{b}}$$simplifying gives$$\sqrt2=\sqrt{\frac{3a}{b}}$$which is simply $2b=3a$.
$2b=3a$ and $ab$ must be square number. However, this is impossible because $ab=\frac{3a^2}{2}$ cannot be square number.
Now let's consider the case where $(6-2b)\sqrt{2a}$ and $2P\sqrt{2b}$ are rational. It easily follows that $\sqrt{2a}$ is rational, and $a=2k^2$ for some nonnegative integer $k$. Then $P=(k+1)(\sqrt{6}-\sqrt{2b})$ and $\sqrt{6}-\sqrt{2b}$ must be rational. It means $b=3$, which we ruled out.

Therefore, $(a, b)=(3, 2)$ is the only pair of integers which makes $\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b}$ a rational number.
