If $$\vec{a}$$ and $$\vec{b}$$ makes an angle $$\cos^{-1}\left(\frac{5}{9}\right)$$ with each other, then for $$|\vec{a} + \vec{b}| = \sqrt{2}|\vec{a} - \vec{b}|$$, $$|\vec{a}| = n|\vec{b}|$$. The integer value of $$n$$ is ____

The angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\cos^{-1}\left(\frac{5}{9}\right)$$ and $$|\vec{a}+\vec{b}| = \sqrt{2}|\vec{a}-\vec{b}|$$.

Let $$\cos\theta = \frac{5}{9}$$.

From the formulas for the square of the length of the sum and difference of vectors we have:

$$|\vec{a}+\vec{b}|^2 = a^2 + b^2 + 2ab\cos\theta$$

$$|\vec{a}-\vec{b}|^2 = a^2 + b^2 - 2ab\cos\theta$$

Using the given condition we write:

$$a^2 + b^2 + 2ab\cos\theta = 2\bigl(a^2 + b^2 - 2ab\cos\theta\bigr)$$

Expanding the right side gives:

$$a^2 + b^2 + 2ab\cos\theta = 2a^2 + 2b^2 - 4ab\cos\theta$$

Collecting like terms leads to:

$$6ab\cos\theta = a^2 + b^2$$

Substituting $$\cos\theta = \frac{5}{9}$$ into the equation gives:

$$6ab \cdot \frac{5}{9} = a^2 + b^2$$

which simplifies to:

$$\frac{10ab}{3} = a^2 + b^2$$

Assume $$|\vec{a}| = n|\vec{b}|$$ so that $$a = nb$$. Substituting into the last equation yields:

$$\frac{10n b^2}{3} = n^2b^2 + b^2$$

Dividing both sides by $$b^2$$ gives:

$$\frac{10n}{3} = n^2 + 1$$

Rearranging to a standard quadratic form:

$$3n^2 - 10n + 3 = 0$$

Factoring the quadratic leads to:

$$ (3n-1)(n-3) = 0$$

Thus $$n = 3$$ or $$n = \frac{1}{3}$$. Since the integer value is required we choose $$n = 3$$.

The answer is $$\boxed{3}$$.