Complex Egyptian Fractions

$\dfrac{5}{4-3i} = \dfrac{5(4+3i)}{(4-3i)(4+3i)} = \dfrac{4+3i}{5}$ .

Let $v = a+bi$ ; $w = c+di$ ; $z = e+fi$ for some integers $a,b,c,d,e,f$ .

Then $\dfrac{1}{v} = \dfrac{a-bi}{|v|^2}$ . $\dfrac{1}{w} = \dfrac{c-di}{|w|^2}$ . $\dfrac{1}{z} = \dfrac{e-fi}{|z|^2}$ .

Hence, $\dfrac{4}{5} = \dfrac{a}{|v|^2} + \dfrac{c}{|w|^2} + \dfrac{e}{|z|^2}$ , and $\dfrac{3}{5} = \dfrac{-b}{|v|^2} - \dfrac{d}{|w|^2} - \dfrac{f}{|z|^2}$ .

With $5$ as a denominator on the left and distinct absolute values on the right, one of these Gaussian integers will have common Gaussian prime factor with $5$ , and the least absolute value will equal to $\sqrt{1^2+2^2} = \sqrt{5}$ .

In other words, one of the Gaussian integer will be one in the set: {1+2i, 1-2i, -1+2i, -1-2i, 2+i, 2-i, -2+i, -2-i}.

Since we are attempting to find the least possible absolute value sum, we can begin by investigate the absolute value $\sqrt{1^2+1^2} = \sqrt{2}$ as the other value. Then, $|v|^2 = 2$ , for all integers' absolute values are greater than $1$ .

Thereby, $v$ will be one in the set: {1+i, 1-i, -1+i, -1-i}.

For the last Gaussian integer, we can opt the square absolute power of $4, 8, 9, 10,...$ as $3, 6, 7$ can not be represented sum of two perfect squares.

$$ Case #1: $|v|^2 = 2$ , $|w|^2 = 4$ ; $|z|^2 = 5$

$\dfrac{4}{5} = \dfrac{a}{2} + \dfrac{c}{4} + \dfrac{e}{5}$ , and $\dfrac{3}{5} = \dfrac{-b}{2} - \dfrac{d}{4} - \dfrac{f}{5}$ .

For $a,b$ , the values vary within $\pm 1$ only. For $c,d$ , one of them is $0$ , and the other is $\pm 2$ . For $e,f$ , one of them is $\pm 1$ , and the other is $\pm 2$ .

If $a = -1$ , the $\dfrac{13}{10} = \dfrac{c}{4} + \dfrac{e}{5}$ . With highest possible values of $c,e$ , it will never exceed $1$ , so $a = 1$ .

Then $\dfrac{3}{10} = \dfrac{c}{4} + \dfrac{e}{5}$ .

Similarly, if $b = 1$ , the equation is also not possible, so $b = -1$ , and $\dfrac{1}{10} = \dfrac{-d}{4} - \dfrac{f}{5}$ .

Since either $c$ or $d$ is zero, that means $\dfrac{3}{10} = \dfrac{e}{5}$ or $\dfrac{1}{10} = \dfrac{f}{5}$ , or $e=1.5$ or $f=0.5$ . However, $e,f$ are integers, so it's contradicted, making case #1 impossible.

Similarly, with one the square absolute value of $9$ , we will reach similar conclusion.

$$

Case #2: $|v|^2 = 2$ , $|w|^2 = 5$ ; $|z|^2 = 8$

$\dfrac{4}{5} = \dfrac{a}{2} + \dfrac{c}{5} + \dfrac{e}{8}$ , and $\dfrac{3}{5} = \dfrac{-b}{2} - \dfrac{d}{5} - \dfrac{f}{8}$ .

For $a,b$ , the values vary within $\pm 1$ only. For $c,d$ , one of them is $\pm 1$ , and the other is $\pm 2$ . For $e,f$ , the values vary within $\pm 2$ only.

That means, $\dfrac{4}{5} = \dfrac{a}{2} + \dfrac{c}{5} \pm \dfrac{1}{4}$ , and $\dfrac{3}{5} = \dfrac{-b}{2} - \dfrac{d}{5} \pm \dfrac{1}{4}$ .

$\dfrac{16}{20} = \dfrac{10a + 4c \pm 5}{20}$ and $\dfrac{12}{20} = \dfrac{-10b -4d \pm 5}{20}$ .

However, the numerators on the right will be odd, so both equations can never be achieved, making case #2 also impossible.

$$

Case #3: $|v|^2 = 2$ , $|w|^2 = 5$ ; $|z|^2 = 10$

$\dfrac{4}{5} = \dfrac{a}{2} + \dfrac{c}{5} + \dfrac{e}{10}$ , and $\dfrac{3}{5} = \dfrac{-b}{2} - \dfrac{d}{5} - \dfrac{f}{10}$ .

For $a,b$ , the values vary within $\pm 1$ only. For $c,d$ , one of them is $\pm 1$ , and the other is $\pm 2$ . For $e,f$ , one of them is $\pm 1$ , and the other is $\pm 3$ .

Then, $\dfrac{8}{10} = \dfrac{5a+2c+e}{10}$ , and $\dfrac{6}{10} = \dfrac{-5b-2d-f}{10}$ .

Hence, there are two possibilities: $a = 1, c=1, e=1, b=-1, d=-2, f=3$ or $a = 1, c=2, e=-1, b=-1, d=1, f=-3$ .

In other words, $\dfrac{5}{4-3i} = \dfrac{1}{1-i} + \dfrac{1}{1-2i} + \dfrac{1}{1+3i} = \dfrac{1}{1-i} + \dfrac{1}{2+i} + \dfrac{1}{-1-3i}$ .

However, only the former case works for $v+w+z = 3$ , which is a real integer.

$$

Finally, since there are no other less value possibility, we can conclude that the least $|v|^2 + |w|^2 + |z|^2 = \boxed{17}$ .