Find the largest possible integer n such that $\sqrt{n}+\sqrt{n+60}=m$ for some non-square integer m. The solution to this problem says that: 
Squaring both sides gives us that $n(n+60)$ is a perfect square
I do not understand this, can someone explain me why this is true.
squareing gives:
$2n+60+2\sqrt{n}\sqrt{n+60}=m^2$
embaresed that i missed it 
$\sqrt{n}\sqrt{n+60}$ must be equal to some positive integer which implies that
$n(n+60)$ must be equal to some squared integer 

As you have found out it is necessary that $n(n+60)$ is a perfect square. Therefore we want $n(n+60)=(n+r)^2$, or $(60-2r)n=r^2$, for some integer $r\geq0$. It follows that $r$ has to be even, and writing $r=2s$ we arrive at the condition $$(15-s)n=s^2\tag{1}$$ for  integer $s$ and $n\geq0$. Considering the cases $0\leq s\leq 15$ in turn we see that the following pairs $(s,n)$ solve $(1)$:
$$(0,0),\quad(6,4),\quad(10,20),\quad(12,48),\quad(14,196)\ .$$
The largest occuring $n$ in this list is $196$, and this $n$ indeed leads to an integer $m=\sqrt{196}+\sqrt{196+60}=30$. As $30$ is not a square the answer to the question is $196$.
