## As we all know the equation $x^2=-1$ has no solutions for real $x$.

If we however introduce the imaginary number $i$ this equation has two solutions: $x=i$ and $x=-i$.

If we go a step further the equation $(x-3)^2=-4$ has two complex solutions: $x=3+2i$ and $x=3-2i$.

$x=3+2i$ and $x=3-2i$ are called each others’ complex conjugate.

Numbers of the form $a+bi$ are called complex numbers.

In general $a+bi$ and $a-bi$ are each other’s complex conjugate.

A Gaussian Integer is a complex number $a+bi$ such that both $a$ and $b$ are integers.

The regular integers are also Gaussian integers (with $b=0$).

To distinguish them from Gaussian integers with $b \ne 0$ we call such integers “rational integers.”

A Gaussian integer $a+bi$ is called a divisor of a rational integer $n$ if the result $\dfrac n {a + bi}$ is also a Gaussian integer.

If for example we divide $5$ by $1+2i$ we can simplify $\dfrac{5}{1 + 2i}$ in the following manner:

Multiply numerator and denominator by the complex conjugate of $1+2i$: $1-2i$.

The result is $\dfrac{5}{1 + 2i} = \dfrac{5}{1 + 2i}\dfrac{1 – 2i}{1 – 2i} = \dfrac{5(1 – 2i)}{1 – (2i)^2} = \dfrac{5(1 – 2i)}{1 – (-4)} = \dfrac{5(1 – 2i)}{5} = 1 – 2i$.

So $1+2i$ is a divisor of $5$.

Note that $1+i$ is not a divisor of $5$ because $\dfrac{5}{1 + i} = \dfrac{5}{2} – \dfrac{5}{2}i$.

Note also that if the Gaussian Integer $(a+bi)$ is a divisor of a rational integer $n$, then its complex conjugate $(a-bi)$ is also a divisor of $n$.

In fact, $5$ has six divisors such that the real part is positive: $\{1, 1 + 2i, 1 – 2i, 2 + i, 2 – i, 5\}$.

The following is a table of all of the divisors for the first five positive rational integers:

$n$ Gaussian integer divisors

with positive real partSum $s(n)$ of these

divisors$1$$1$$1$

$2$$1, 1+i, 1-i, 2$$5$

$3$$1, 3$$4$

$4$$1, 1+i, 1-i, 2, 2+2i, 2-2i,4$$13$

$5$$1, 1+2i, 1-2i, 2+i, 2-i, 5$$12$

For divisors with positive real parts, then, we have: $\sum \limits_{n = 1}^{5} {s(n)} = 35$.

$\sum \limits_{n = 1}^{10^5} {s(n)} = 17924657155$.

What is $\sum \limits_{n = 1}^{10^8} {s(n)}$?

### The problem at hand involves summation of divisors of a number. Theoretically, you could compute the sum for each individual number from 1 to 10^8, but the sheer size of the numbers we’re dealing with make this approach impractical. If you were dealing with normal integers, you could use well-known mathematical tools like the divisor function to simplify this process. However, since we’re dealing with Gaussian integers, things are a bit more tricky.

This is beyond the capability of a human tutor to calculate directly, you would need to write or use a pre-existing program to do so. Here is a general approach one might take:

1. Create an array with size 10^8, initialized to 0.

2. Loop over each Gaussian integer a + bi where a, b ranging from -10^4 to 10^4 and update the corresponding position in array as multiples of the Gaussian integer (considering only those have a positive real part). Perform this operation carefully as it will have to deal with the overflow situation (since the numbers are very large).

3. Finally, perform summing operation over the array to get the desired result.

This algorithm is highly parallelizable and you should leverage this as much as you can, using either multicore programming (in a language that supports it) or leveraging a platform that supports distributed computing.

Remember, the calculation also requires significant computation resources and may take considerable time to execute regardless of the method used. It’s a problem that’s meant to test computational methods more than it is to test mathematical comprehension.

In conclusion, while the math understanding needed here is basic (such as understanding how complex numbers are), this type of problem actually falls under the realm of computer science and numerical problem solving, much more than pure mathematics.

##### More Answers:

Sub-triangle SumsA Preference for A5

Sums of Square Reciprocals