4. Solve the equation

$$
\frac{1}{x}+\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}-\frac{1}{x+8}-\frac{1}{x+10}+\frac{1}{x+12}+\frac{1}{x+14}=0
$$

S o l u t i o n. Substituting $\mathrm{y}=\mathrm{x}+7$ makes the given equation symmetric:

$$
\frac{1}{y-7}+\frac{1}{y-5}-\frac{1}{y-3}-\frac{1}{y-1}-\frac{1}{y+1}-\frac{1}{y+3}+\frac{1}{y+5}+\frac{1}{y+7}=0
$$

Grouping as follows

$$
\left(\frac{1}{y-7}+\frac{1}{y+7}\right)+\left(\frac{1}{y-5}+\frac{1}{y+5}\right)-\left(\frac{1}{y-3}+\frac{1}{y+3}\right)-\left(\frac{1}{y-1}+\frac{1}{y+1}\right)=0
$$

This gives

$$
\frac{2 y}{y^{2}-49}+\frac{2 y}{y^{2}-25}-\frac{2 y}{y^{2}-9}-\frac{2 y}{y^{2}-1}=0
$$

i.e.

$$
2 y\left(\frac{1}{y^{2}-49}+\frac{1}{y^{2}-25}-\frac{1}{y^{2}-9}-\frac{1}{y^{2}-1}\right)=0
$$

from which $y=0$, i.e. $x=-7$, or (after substituting $z=y^{2}$ )

$$
\frac{1}{z-49}+\frac{1}{z-25}-\frac{1}{z-9}-\frac{1}{z-1}=0
$$

Grouping

$$
\left(\frac{1}{z-49}-\frac{1}{z-1}\right)+\left(\frac{1}{z-25}-\frac{1}{z-9}\right)=0
$$

This gives

$$
\frac{48}{z^{2}-50 z+49}+\frac{16}{z^{2}-34 z+225}=0
$$

Simplifying by 16 and bringing to a common denominator, we get

$$
3\left(z^{2}-34 z+225\right)+\left(z^{2}-50 z+49\right)=0
$$

and, of course, $\mathbf{z} \neq 1, \mathbf{z} \neq 9, \mathbf{z} \neq 25, \mathbf{z} \neq 49$. Combining like terms, we have $4 \cdot z^{2}-152 z+724=0$, from which, simplifying by 4, we get $z^{2}-38 z+181=0$. The roots of this equation are

$$
z=y^{2}=19 \pm 6 \sqrt{5},
$$

from which $y= \pm \sqrt{19 \pm 6 \sqrt{5}}$ and, finally, $x=y-7=-7 \pm \sqrt{19 \pm 6 \sqrt{5}}$, with various combinations of signs possible.
Answer: $\mathrm{x}=-7, \mathrm{x}=-7 \pm \sqrt{19 \pm 6 \sqrt{5}}$.