Solve the following equation:

$$
16 x(x+1)(x+2)(x+3)=9
$$
# I. Solution.

$$
\begin{gathered}
x(x+3)=x^{2}+3 x \\
(x+1)(x+2)=x^{2}+3 x+2
\end{gathered}
$$

If we now let $x^{2}+3 x=y$, then our equation becomes:

$$
16 y(y+2)=9 \quad \text { or } \quad 16 y^{2}+32 y-9=0
$$

From this, $y=\frac{-32 \pm \sqrt{32^{2}+4 \cdot 16 \cdot 9}}{32}=\frac{-32 \pm 40}{32} ; \quad y_{1}=\frac{-9}{4}, \quad y_{2}=+\frac{1}{4}$.

I. If $\quad x^{2}+3 x=-\frac{9}{4}, \quad$ then $\quad x=\frac{-3 \pm \sqrt{9-9}}{2}=\frac{-3}{2}$.

II. If $\quad x^{2}+3 x=+\frac{1}{4} . \quad, \quad x=\frac{-3 \pm \sqrt{9+1}}{2}=\frac{-3+\sqrt{10}}{2}$.

Accordingly, the original fourth-degree equation has four roots:

$$
\frac{-3-\sqrt{10}}{2}, \quad-\frac{3}{2}, \quad-\frac{3}{2}, \quad \frac{-3+\sqrt{10}}{2}
$$

$\left(\frac{3}{2}\right.$ is a double root of the equation.)

László Irányi (Kegyesrendi Gymnasium, 6th grade, Szeged)

II. Solution. In addition to the findings introduced in the previous solution,

$$
\begin{gathered}
x(x+1)(x+2)(x+3)=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)=\left(x^{2}+3 x\right)^{2}+2\left(x^{2}+3 x\right)= \\
\left(x^{2}+3 x+1\right)^{2}-1
\end{gathered}
$$

If $x^{2}+3 x+1=z$, then

(II.)

$$
\begin{array}{rcc}
16\left(z^{2}-1\right)=9, \quad z^{2}-1=\frac{9}{16}, & z^{2}=\frac{25}{16}, \quad z= \pm \frac{5}{4} \\
x^{2}+3 x+1=\frac{5}{4} \quad \text { or } & x^{2}+3 x=\frac{1}{4} \\
x^{2}+3 x+1=-\frac{5}{4} & \text { or } & x^{2}+3 x=-\frac{9}{4}
\end{array}
$$

The solutions to these two equations are found in I.

Note. In the 1st issue of the 3rd year, in exercise 133, we proved that if we add 1 to the product of four consecutive integers, we get a perfect square. This can be generalized:

if we add the fourth power of the common difference to the product of four consecutive terms of an arithmetic progression, we get a perfect square expression. Indeed,

$$
\begin{gathered}
x(x+d)(x+2 d)(x+3 d)+d^{4}=\left(x^{2}+3 d x\right)\left(x^{2}+3 d x+2 d^{2}\right)+d^{4}= \\
=\left(x^{2}+3 d x\right)^{2}+2\left(x^{2}+3 d x\right) d^{2}+d^{4}=\left(x^{2}+3 d x+d^{2}\right)^{2}
\end{gathered}
$$

In our case, $d=1$.