If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$ 
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$

This is a 1989 ARML problem. One, ugly way to solve this is:

Approximate this as $x^4=379^2$, so $x\approx \sqrt{379}\approx 19$ and guess and check around there to see that $18$ works.

What's a nicer way? 
Hint

 Difference of squares


Note that
\begin{align*}
x(x+1)(x+2)(x+3) &=379^2-1\\
&=(380)(378) \\
&=(19)(20)(18)(21).
\end{align*}
Hence it follows that $x=18$.
