Inverse Tangent Difference as a Single Fraction

The general term of the series is:

$$T_n = \tan^{-1}\left(\frac{2n}{n^4 + n^2 + 2}\right)$$

We can rewrite the denominator to expose the standard form $$1 + ab$$:

$$n^4 + n^2 + 2 = 1 + (n^4 + n^2 + 1)$$

Using the algebraic factorization $$(n^4 + n^2 + 1) = (n^2 + n + 1)(n^2 - n + 1)$$, the term becomes:

$$T_n = \tan^{-1}\left(\frac{2n}{1 + (n^2 + n + 1)(n^2 - n + 1)}\right)$$

We can rewrite the numerator $$2n$$ as the difference of these two factors:

$$2n = (n^2 + n + 1) - (n^2 - n + 1)$$

Substituting this into the expression for $$T_n$$:

$$T_n = \tan^{-1}\left(\frac{(n^2 + n + 1) - (n^2 - n + 1)}{1 + (n^2 + n + 1)(n^2 - n + 1)}\right)$$

Using the identity $$\tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1}a - \tan^{-1}b$$, we get:

$$T_n = \tan^{-1}(n^2 + n + 1) - \tan^{-1}(n^2 - n + 1)$$

Let $$V_n = \tan^{-1}(n^2 + n + 1)$$. Notice that:

$$V_{n-1} = \tan^{-1}((n-1)^2 + (n-1) + 1) = \tan^{-1}(n^2 - n + 1)$$

Thus, the general term telescopes as:

$$T_n = V_n - V_{n-1}$$

The partial sum $$S_k$$ up to $$k$$ terms is:

$$S_k = \sum_{n=1}^{k} (V_n - V_{n-1}) = V_k - V_0$$

Substituting the definitions back in:

$$S_k = \tan^{-1}(k^2 + k + 1) - \tan^{-1}(0^2 + 0 + 1)$$

$$S_k = \tan^{-1}(k^2 + k + 1) - \tan^{-1}(1)$$

Taking the limit as $$k \to \infty$$ for the infinite sum:

$$S = \lim_{k \to \infty} \left[\tan^{-1}(k^2 + k + 1) - \tan^{-1}(1)\right]$$

$$S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$

Using the identity

$$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$$

(valid after checking the principal values), the left-hand side becomes

$$\tan^{-1}\left(\frac{\frac{x}{1+2x^2}+\frac{x}{1+6x^2}}{1-\frac{x^2}{(1+2x^2)(1+6x^2)}}\right)$$

Simplifying,

$$\frac{\frac{x(2+8x^2)}{(1+2x^2)(1+6x^2)}}{\frac{1+7x^2+12x^4-x^2}{(1+2x^2)(1+6x^2)}}=\frac{2x(1+4x^2)}{1+6x^2+12x^4}$$

Hence the left-hand side is

$$\tan^{-1}\left(\frac{2x(1+4x^2)}{1+6x^2+12x^4}\right)$$

Now for the right-hand side,

$$\tan^{-1}(2x)-\tan^{-1}\left(\frac{x}{2}\right)=\tan^{-1}\left(\frac{2x-\frac{x}{2}}{1+x^2}\right)$$

$$=\tan^{-1}\left(\frac{3x}{2(1+x^2)}\right)$$

Since both arguments lie in the principal range, equate the arguments:

$$\frac{2x(1+4x^2)}{1+6x^2+12x^4}=\frac{3x}{2(1+x^2)}$$

Cross-multiplying,

$$4x(1+4x^2)(1+x^2)=3x(1+6x^2+12x^4)$$

$$x\Big(4(1+5x^2+4x^4)-3(1+6x^2+12x^4)\Big)=0$$

$$x(1+2x^2-20x^4)=0$$

Thus,

$$x=0$$

or

$$20x^4-2x^2-1=0$$

Let

$$y=x^2$$

Then

$$20y^2-2y-1=0$$

$$y=\frac{2\pm\sqrt{84}}{40}=\frac{1\pm\sqrt{21}}{20}$$

Only

$$y=\frac{1+\sqrt{21}}{20}>0$$

is admissible.

Hence

$$x=\pm\sqrt{\frac{1+\sqrt{21}}{20}}$$

Together with

$$x=0$$

there are

$$3$$

real solutions.

Therefore, $$\text{Answer}=3$$.