---
id: 5900f4051000cf542c50ff18
title: 'Problem 153: Investigating Gaussian Integers'
challengeType: 5
forumTopicId: 301784
dashedName: problem-153-investigating-gaussian-integers
---

# --description--

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 $x = 3 - 2i$, which 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 $b ≠ 0$ we call such integers "rational integers."

A Gaussian integer is called a divisor of a rational integer $n$ if the result is also a Gaussian integer.

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

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

The result is:

$$\frac{5}{1 + 2i} = \frac{5}{1 + 2i} \frac{1 - 2i}{1 - 2i} = \frac{5(1 - 2i)}{1 - {(2i)}^2} = \frac{5(1 - 2i)}{1 - (-4)} = \frac{5(1 - 2i)}{5} = 1 - 2i$$

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

Note that $1 + i$ is not a divisor of 5 because:

$$\frac{5}{1 + i} = \frac{5}{2} - \frac{5}{2}i$$

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 part | Sum 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: $\displaystyle\sum_{n=1}^5 s(n) = 35$.

For $1 ≤ n ≤ {10}^5$, $\displaystyle\sum_{n = 1}^{{10}^5} s(n) = 17924657155$.

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

# --hints--

`sumGaussianIntegers()` should return `17971254122360636`.

```js
assert.strictEqual(sumGaussianIntegers(), 17971254122360636);
```

# --seed--

## --seed-contents--

```js
function sumGaussianIntegers() {

  return true;
}

sumGaussianIntegers();
```

# --solutions--

```js
// solution required
```