<s>[INST] 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

 [/INST] $$x(x+1)(x+2)(x+3) +1 =(x^2 +x)(x^2 +5x +6) +1 = x^4+5x^3 +6x^2 +x^3 +5x^2 +6x +1 =x^4 +6x^3 +11x^2 +6x +1 =x^2 (x^2 +6x +9) +2x(x+3)+1 =(x(x+3) +1)^2$$
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