Find all natural roots of $\sqrt{x}+\sqrt{y}=\sqrt{1376}$ given that $x\leq y$ 
Find all natural roots of $\sqrt{x}+\sqrt{y}=\sqrt{1376}$ given that $x\leq y$

I'm confused of this equation because $1376$ is not a square!!
So maybe it has no natural root! Am I right??
 A: Write it as:
$\sqrt{y}=\sqrt{1376}-\sqrt{x}$
hence:
$y=1376-2\sqrt{1376x}+x$
$y=1376-8\sqrt{86x}+x$
Hence $x$ must be a multiple of $86$ as $86x$ must be a square number.
Let $x=86n^2$
$y=1376-688n+86n^2$
From the original problem $0\leq y\leq 1376$ so:
$0\leq1376-688n+86n^2$ and $1376-688n+86n^2\leq1376$
The second equation gives $0\leq n\leq8$. Whilst the first equation gives reduces to $0\leq86(n-4)^2$ which tells us that the solutions are symmetrical around $n=4$ so we only need consider $n=0\cdots4$.
So consider $n=0\cdots4$ which corresponds to $(x,y)=(0,1376),(86,774),(344,344),(774,86),(1376,0)$. The last ones can be dropped due to $x\leq y$ hence:
$(x,y)=(0,1376),(86,774),(344,344)$.
A: Square the equation:
$x+y+2\sqrt{xy}=1376$ or $2\sqrt{xy}=1376-x-y$
Square again:
$4xy=(1376-x-y)^2$
Solve the resulting quadratic equation.

Additionally you could try to solve the simplified equation $x=y$
$2\sqrt{y}=\sqrt{1376}$
$2\sqrt{y}=2\sqrt{4 \cdot 2 \cdot 43}$
$y=4\cdot 2 \cdot 43$
A: W.L.O.G. suppose that $x \ge y$. Let $x=ad$ and $y=bd$ where $\gcd(a, b)=1$. So$$a+b+2\sqrt{ab}=\frac{1376}{d}$$Thus $ab$ must be a perfect square and since $a$ and $b$ are co-prime both of them must be squares. Let $a=A^2$ and $b=B^2$, equation becomes$$(A+B)^2=\frac{1376}{d}$$Thus $1376/d>1$ is a perfect square. But notice that $1376=2^5 \times 43$, hence $1376/d= 4, 16$, which leads to $(A, B)=(1, 1)$ and $(A,B)=(3, 1)$ and to $(x, y)=(344, 344)$ and $(x, y)=(774, 86)$, respectively.   
