Squaring Not Help (maybe..)

Okay i think its pretty obvious. If you factor $2014$ , it gives $2,19,53$

And none of its factors is a perfect square, all of them are primes. Which implies that there will be $0$ integral solutions.

Let me explain it with an example. Lets take the number $27$ in place of $2014$ We have $\sqrt{m} + \sqrt{n} = \sqrt{27}$

$\sqrt{m} + \sqrt{n} = \sqrt{9.3}$ we can split $27$ into $9.3$ and here $9$ is a perfect square

$\sqrt{m} + \sqrt{n} = 3 \sqrt{3} \$ and now its easy to see that one term on $LHS$ will be $2 \sqrt{3}$ and the other will be $\sqrt{3}$

But this is not possible with $2014$ because it has only 3 prime factors, we can't covert their products to same radicals and add them up, its not possible.

Moreover, $m$ and $n$ should have the same prime factors because if not, then we can't add them up and receive one single radical.

Note that for m,n to be positive integers

$m < 2014$ and $n < 2014$

Rearranging the original equation :

$\sqrt{m} = \sqrt{2014} - \sqrt{n}$

Squaring we get :

$m = 2014 + n - 2*\sqrt{2014*n}$

Prime factorization of $2014 = 2*19*53$ , hence :

$m = 2014 + n - 2*\sqrt{2*19*53*n}$

Thus for this equation to have positive integral solution , $(2*19*53*n)$ must be a perfect square which is only possible if :

$n = 2*19*53$

$n = 2014$

which is not possible as $n < 2014$ .

Hence the given equation has no solutions.

$\textbf{Note :}$ By the same argument we can show that this kind of an equation ;

$\sqrt{m} + \sqrt{n} = \sqrt{k}$

has no solutions if k does not contain a prime factor with power atleast 2.