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

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.

• How about $x = 0$, and $y = 135$? – user49685 Feb 11 '14 at 18:44
• I think you need some extra info here. – John Habert Feb 11 '14 at 18:45
• So what you want to do is find the sum of x and y, when you know the square root of x plus the square root of y equals the square root of 135. – user2608474 Feb 11 '14 at 18:49
• There are not many integer solutions $(x,y)$, but there are infinitely many real solutions. – André Nicolas Feb 11 '14 at 18:50
• That $x$ and $y$ are positive integers is the extra info needed. Thanks. – John Habert Feb 11 '14 at 18:53

## 3 Answers

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.

• @user49685 fixed it, ty. – N. S. Feb 11 '14 at 18:58
• I am afraid but that's not the answer, unfortunately. – user97615 Mar 23 '15 at 21:44

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 ...

• Hint: You can rewrite your galois theory in terms of rationalization: $\sqrt{15x}-\sqrt{15y}=\frac{45}{\sqrt{15x}+\sqrt{15y}}$ is rational, thus both roots are rational, thus integer....Or that way :) – N. S. Feb 11 '14 at 18:57

As a rule, for all $x,y,z \in \mathbb{Z}>0$ , such that $\sqrt{z}=\sqrt{x}+\sqrt{y}$, we have the following:

Set $z=h^2u$ where $h,u \in \mathbb{Z}>0$ ($u$ is square-free) then $\sqrt{z}=h\sqrt{u}$

We can rewrite: $$\sqrt{z}=(h-g+g)\sqrt{u}$$ with $g$ is any integer. $$h\sqrt{u}=(h-g)\sqrt{u}+g\sqrt{u}$$ After putting the coefficients of $\sqrt{u}$ under the radicands, we obtain:

$$h\sqrt{u}=\sqrt{(h-g)^2u}+\sqrt{g^2u}$$So now it's easy to get the solutions to your problem:$$x=\sqrt{(h-g)^2u}$$

$$y=\sqrt{g^2u}$$$$z=135=3^2(15)$$ which implies $h=3$ and $u=15$

We can also deduct from $h=3$ that $(h-g)$ or $(g)$ is equal to either $1$ or $2$ and versa versa. The case where $g=0$ is clearly trivial.

WLOG, we set: $$h-g=1$$ since $h=3$ then $$g=2$$

As a result, we have:$$x+y={(h-g)^2u}+{g^2u}$$ by replacing $u=15$, $(h-g)=1$ and $g=2$

We obtain$$x+y=15(1)^2+2^2(15)=75$$ QED