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

 A: 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 $x=18$.
A: Outline
$x(x+1)(x+2)(x+3) + 1 = (x^2 + 3x + 1)^2 = 379^{2}$
$(x - 18)(x + 21) = 0$
$\color{blue}{x = 18}$
A: 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 $x$.
A: 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 $x=n-1$.
A: $$x(x+1)(x+2)(x+3) +1 =(x^2 +x)(x^2 +5x +6) +1 = x^4+5x^3 +6x^2 +x^3 +5x^2 +6x +1 =x^4 +6x^3 +11x^2 +6x +1 =x^2 (x^2 +6x +9) +2x(x+3)+1 =(x(x+3) +1)^2$$
