Prove $\sqrt{n}+\sqrt{n+k}$ is irrational For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)

This result is not true in general for example, suppose that $n$ is a square $n=l^2$ and $k=0$ $\sqrt{l^2+0}+\sqrt{l^2}$ is rational, suppose that $n=9, k=16$, $\sqrt{16+9}+\sqrt{9}$ is rational.
The question is given $n$ for what values of $k$, $\sqrt{n+k}+\sqrt{n}$ is irrrational?
Proposition
Suppose that $k,n\in N$ and $n(n+k)$ is not a square  then $\sqrt{n+k}+\sqrt{n}$ is irrational.
Proof:
$(\sqrt{n}+\sqrt{n+k})^2=n+n+k+2\sqrt{n}\sqrt{n+k}$
$((\sqrt{n}+\sqrt{n+k})^2-2n-k)^2=4n(n+k)$
Consider $P(X)= (X^2-2n-k)^2-4n(n+k)=X^4-2(2n+k)X^2+(2n+k)^2-4n(n+k)=X^4-2(2n+k)X^2+k^2$ $\sqrt{n+k}+\sqrt{n}$ is a root of $P(X)$. 
Consider $Q(X)=U^2-2(2n+k)U+k^2$. The discriminant of $Q(X)$ is $4(2n+k)^2-4k^2=4(4n^2+4nk+k^2)-4k^2 =16n(n+k)$, thus if $n(n+k)$ is not a square, $Q(X)$ and hence $P(X)$ is irreducible.
