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 

Obviously both $n$ and $n+60$ should be  squares so its product $n(n+60)$ is a square. On the other hand, $n$ cannot be greater than some integer because if $n=x^2$ and $n+60=y^2$ then making $y=x+h$ we must have $2xh+h^2= 60$ which clearly put a bound for $x$.
This bound is easy to calculate and one has $x\lt 30$ because $30^2+60\lt 31^2$ hence $30^2+60\lt 31^2$ cannot be a square.
Furthermore $$\begin{cases}n=x^2\\n+60=y^2\end{cases}\Rightarrow(y+x)(y-x)=60=2^2\cdot3\cdot5\Rightarrow(x,y)=(14,16)$$ this largest $(14,16)$ corresponding to the factorization $30\cdot2$ (the factorization $60\cdot1$ does not give integer solution).
Thus the largest possible $n$ is $\color{red}{n=14^2=196}$ which corresponds to the integer $m=30=14+16$
