3.0 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f4051000cf542c50ff18 | Problem 153: Investigating Gaussian Integers | 5 | 301784 | 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
.
assert.strictEqual(sumGaussianIntegers(), 17971254122360636);
--seed--
--seed-contents--
function sumGaussianIntegers() {
return true;
}
sumGaussianIntegers();
--solutions--
// solution required