Prove $\sqrt{n}+\sqrt{n+k}$ is irrational For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)
 A: Well, a possible but perhaps in the long run not exhaustive method, since you must find for what $k \in \mathbb{N}$ this is true; consider
$$(\sqrt{n}+\sqrt{n+k})(-\sqrt{n}+\sqrt{n+k})=-n+(n+k)$$
Which of course gives $k$.
now, since $k$ is rational, it means that $\sqrt{n}$ and $\sqrt{n+k}$ is rational thus
\begin{align}
n&=&p^{2} \\
n+k&=& q^{2}
\end{align}
Perhaps you can see where I am going with this and finish it off?
A: Hint:
If $k=2m+1$, $m\ge1$, then $\sqrt{n}+\sqrt{n+k}$ fails to be an irrational number for $n=m^2$:
$$\sqrt{n}+\sqrt{n+k}=\sqrt{m^2}+\sqrt{m^2+2m+1}=m+(m+1)\in\mathbb{Q}$$
If $k=4m$, $m>1$, then $\sqrt{n}+\sqrt{n+k}$ fails to be an irrational number for $n=(m-1)^2$.
$$\sqrt{n}+\sqrt{n+k}=\sqrt{(m-1)^2}+\sqrt{(m-1)^2+4m}=(m-1)+(m+1)\in\mathbb{Q}$$
A: 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.
A: Solution is $k = 2^{2v+1}*j$ where j is odd.
Obviously $k$ cannot equal $0$.
$k$ must/can be an odd number multiplied by an odd power of 2. ($k$ can not be odd.  If $k$ is even, $k$ is not an odd number times an even power of 2.
If $0$ is not considered a natural number than 1 is a possible value for $k$ is the one exception.  (As $\sqrt{n} + \sqrt{n+1}$ is rational $\iff n = 0$).
$\sqrt{n} + \sqrt{n + k} = r = a/b; \gcd(a,b) = 1 \implies n + k = r^2 + n - 2r\sqrt{n} \implies k = r(r - 2\sqrt{n}) \implies k = (a/b)(a/b - 2m); n = m^2$ for some integer $m$.  Which implies $bk = a^2/b - 2am \in \mathbb Z \implies b = 1$.
So for any $k = a(a - 2m)$ we have a possibility of the sum being rational.  Otherwise we don't.
So we can't have $k = 2j + 1$ odd or we'd have $k = k(k - 2j)$ (which would yield $\sqrt{j^2} + \sqrt{j^2 + k = j^2 + 2j + 1}$ being rational.)
We can't have $k = 2^{2v}j$ where $j$ is odd or we'd have $k = 2^vj(2^vj - 2^v(j - 1))$
On the other hand if $k = 2^{2v + 1}j$ where $j$ is odd we can't have $k = a(a - 2m)$.  The powers of 2 just don't add up.
A: Suppose that $$\sqrt n +\sqrt {n+k}=r, \quad r\in\mathbb{Q}$$
Now we have: $$\sqrt {n+k} =r-\sqrt n $$
$$n+k=r^2-2r\sqrt n +n$$ 
$$2r\sqrt n=r^2-k$$
$$\sqrt n=\frac{r^2-k}{2p}$$
This is a contradiction because the left is $\sqrt n\in\mathbb{I=\mathbb{Q^c}}$, and the right is $\frac{r^2-k}{2p}\in\mathbb{Q}$. The contradiction is due to a wrong assumption that $\sqrt n+\sqrt {n+k}$ was a rational number.
