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 

We need $n(n+60)$ to be a perfect square.  Note that
$$
n(n+60) = (n+30)^2-900
$$
and $900 = 449+451$, so we can have $n+30 = \frac{451}{2} = 226$, or $n = 196$, and $n(n+60) = 50176 = 224^2$.
