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

Question

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

Answer 1

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 $x=18$.

Answer 2

Outline

$x(x+1)(x+2)(x+3) + 1 = (x^2 + 3x + 1)^2 = 379^{2}$

$(x - 18)(x + 21) = 0$

$\color{blue}{x = 18}$

Answer 3

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 $x$.

Answer 4

$$x(x+1)(x+2)(x+3) +1 =(x^2 +x)(x^2 +5x +6) +1 = x^4+5x^3 +6x^2 +x^3 +5x^2 +6x +1 =x^4 +6x^3 +11x^2 +6x +1 =x^2 (x^2 +6x +9) +2x(x+3)+1 =(x(x+3) +1)^2$$