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


About general solution of this equation. 
$x(x+1)(x+2)(x+3)=a^2-1$ 
$⇔(x^2+3x)(x^2+3x+2)=(a-1)(a+1) $
obviously
$x^2+3x=a-1 $ so,
$x=\dfrac{-3+\sqrt{5+4a}}{2}$
$a_n=1,5,11,19・・・
    =1+\sum2(n+1) 
    =1+n(n-1)+2(n-1)=n^2+n-1 $
Therefore when $a=n^2+n-1$, this equation has  one positive solution $x=n-1$.
