Let $$H_n: \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$$, $$n \in \mathbb{N}$$. Let $$k$$ be the smallest even value of $$n$$ such that the eccentricity of $$H_k$$ is a rational number. If $$l$$ is the length of the latus rectum of $$H_k$$, then $$21l$$ is equal to _______

We need to find the smallest even natural number $$n$$ such that the eccentricity of the hyperbola $$H_n: \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$$ is rational, and then compute $$21l$$ where $$l$$ is the length of the latus rectum.

First, we find the eccentricity in terms of $$n$$.

For a hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, the eccentricity is $$e = \sqrt{1 + \frac{b^2}{a^2}}$$.

Here $$a^2 = 1+n$$ and $$b^2 = 3+n$$, so:

$$ e^2 = 1 + \frac{3+n}{1+n} = \frac{(1+n) + (3+n)}{1+n} = \frac{4+2n}{1+n} $$

Next, we determine when $$e$$ is rational.

Let $$f = n + 1$$. Then $$e^2 = \frac{2f + 2}{f} = 2 + \frac{2}{f}$$.

For $$e$$ to be rational, $$e^2$$ must be the square of a rational number. Let $$e = \frac{p}{q}$$ in lowest terms. Then $$\frac{p^2}{q^2} = 2 + \frac{2}{f}$$, which gives $$f = \frac{2q^2}{p^2 - 2q^2}$$.

For $$f$$ to be a positive integer, we need $$p^2 - 2q^2 > 0$$ and $$p^2 - 2q^2$$ must divide $$2q^2$$.

From this, we search for the smallest even $$n$$.

We need $$n$$ even, so $$f = n + 1$$ must be odd. Testing systematically:

Try $$p = 10, q = 7$$: $$p^2 - 2q^2 = 100 - 98 = 2$$, and $$f = \frac{2 \times 49}{2} = 49$$ (odd). So $$n = 48$$ (even).

We should verify no smaller even $$n$$ works. For smaller values of $$f$$ (odd), we need $$2 + 2/f$$ to be a perfect square of a rational. Checking $$f = 1$$ ($$n=0$$, not natural), $$f = 3$$ ($$n=2$$): $$e^2 = 8/3$$, not a perfect square. $$f = 5$$ ($$n=4$$): $$e^2 = 12/5$$, not rational square. Continuing this way, $$f = 7, 9, 11, \ldots, 47$$ all fail to give a rational $$e$$. Thus $$n = 48$$ is the smallest even value.

Now, we verify $$e$$ is rational for $$n = 48$$.

$$ e^2 = \frac{4 + 96}{49} = \frac{100}{49} \implies e = \frac{10}{7} \quad \text{(rational)} \checkmark $$

Then, we calculate the latus rectum length.

For $$H_{48}$$: $$a^2 = 49$$, $$b^2 = 51$$, $$a = 7$$.

The length of the latus rectum of a hyperbola is $$l = \frac{2b^2}{a}$$:

$$ l = \frac{2 \times 51}{7} = \frac{102}{7} $$

Continuing, we compute $$21l$$.

$$ 21l = 21 \times \frac{102}{7} = 3 \times 102 = 306 $$

The answer is 306.