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


Multiply first and last term and middle terms and take 1 on RHS. $(x^2+3x)(x^2+3x+2)=379^2-1^2$ so substitute $x^2+3x=y$ you will get a simple quadratic ie $y(y+2)=380\times 378$ which are also the factors . Get the value of $y$ and and then resubstituting $y=x^2+3x$ you will get the value of $x$ Hope you can take it from here to find $x$.
