How to find the sum of positive integers $x$ and $y$, given that $ \sqrt x + \sqrt y = \sqrt {135} $?

Question

How do you find the sum of integers $x$ and $y$ from: $ \sqrt x + \sqrt y = \sqrt {135} $?

Is there a specific method that will get the answer? x and y are both positive integers. For example x could not be 1, and y could not be 1, because there roots added do not equal the square root of 135. So the sum of x and y could not be 2.

What I thought of doing was kind of like an approximation. Where we know that the sqrt of $135$ is between $11$ and $12$. So we find $2$ numbers that add up to $11$, and then square them, and we get an approximate answer for the sum of $x$ and $y$. So for example, $8$ and $3$. Square them and get $64$ and $9$. We get $73$ for the sum of $x$ and $y$ (the actual answer is $75$).

This method isn't good, so I was wondering if there was another way of doing it.

Answer 1

Squaring both sides you get

$$x+y=135-2 \sqrt{xy}$$

Now, by AM-GM

$$0 \leq 2 \sqrt{xy} \leq \frac{(\sqrt{x}+\sqrt{y})^2}{2}=\frac{135}{2}$$

which tells us that

$$0 \leq x+y \leq 135 \,.$$

And any real number in this range is actually achievable.

If you know more that $x,y$ are integers, then you get more restrictions.

Answer 2

Under the additional assumption that $x,y$ should be integers, note that we can multiply with $\sqrt{15}$ to get $\sqrt{15x}+\sqrt{15y}=45$. The sum of square roots can only be an integer if trivially so (i.e. if we in fact take square roots of perfect squares): $$ \sqrt a+\sqrt b=c\implies a=(c-\sqrt b)^2=c^2+b-2c\sqrt b\implies \sqrt b=\frac{c^2+b-a}{2c}\in\mathbb Q$$ So $15x$ and $15y$ must be perfect squares ...