Problem 153
Investigating Gaussian Integers
As we all know the equation x2=-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 ≠ 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 $\frac{5}{1+2i}$ 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: $\sum_{n=1}^{5}s(n)=35$.
For 1 ≤ n ≤ 105, ∑ s(n)=17924657155.
What is ∑ s(n) for 1 ≤ n ≤ 108?
高斯整数的研究
我们都知道方程x2=-1在实数范围内无解。
但如果我们引入虚数i，这个方程将会有两个解x=i和x=-i。
进一步地，方程(x-3)2=-4有两个复数解：x=3+2i和x=3-2i。
x=3+2i和x=3-2i互称为共轭复数。
形如a+bi的数被称为复数。
概括地说，a+bi和a−bi互称为共轭复数。
高斯整数是形如a+bi且a和b均为整数的复数。
一般意义上的整数也是高斯整数（取b=0）。
为了把它们和b ≠ 0的高斯整数区分开来，称它们为“有理整数”。
如果一个高斯整数除有理整数n的结果仍然是高斯整数，则称它为该有理整数的约数。
例如，我们用1+2i除5，按如下方式简化$\frac{5}{1+2i}$：
分子和分母同时乘以1+2i的共轭：1−2i。
结果是：$\frac{5}{1+2i}=\frac{5}{1+2i}\frac{1-2i}{1-2i}=\frac{5(1-2i)}{1-(2i)^2}=\frac{5(1-2i)}{1-(-4)}=\frac{5(1-2i)}{5}=1-2i$。
所以1+2i是5的约数。
注意1+i不是5的约数，因为$\frac{5}{1+i}=\frac{5}{2}-\frac{5}{2}i$。
同时注意如果高斯整数(a+bi)是有理整数n的约数，那么它的共轭复数(a−bi)也是n的约数。
事实上，5一共有六个实数部分是正数的约数：{1, 1 + 2i, 1 − 2i, 2 + i, 2 − i, 5}。
如下表格列出了前五个正有理整数的所有约数：
| n | 实数部分是正数的高斯整数约数 | 约数的和s(n) |
|---|---|---|
| 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 |
对于实数部分为正数的约数，我们有：$\sum_{n=1}^{5}s(n)=35$。
对于1 ≤ n ≤ 105，∑ s(n)=17924657155。
对于1 ≤ n ≤ 108，求∑ s(n)。