Question

Show that the sum of the squares of all the numbers less than a given number $N$ and prime to it is
$\frac{N^3}{3}(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})....+\frac{V}{6}(1-a)(1-b)(1-c)....,$ and the sum of the cubes is
$\frac{N^4}{4}(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})....+\frac{N^2}{4}(1-a)(1-b)(1-c)...,a,b,c....$ being the difference prime factors of $N$.

Answer 1

$\Rightarrow$ The numbers less than $N$ and not prime to it are given by

$\sum \frac{N}{a}-\sum \frac{N}{ab}+\sum \frac{N}{abc}-.......$

Let us first find the sum of the squares of all numbers less than $N$ and not prime to it

These are given bby the sum of

$a^2+{\left(2a\right)}^2+{\left(3a\right)}^2+......+{(\frac{N}{a}.a)}^2$

$b^2+{\left(2b\right)}^2+{\left(3b\right)}^2+......+{(\frac{N}{b}.b)}^2$

....................................

$-{\left(ab\right)}^2-{\left(2ab\right)}^2-{\left(3ab\right)}^2-......-{(\frac{N}{ab}.ab)}^2$

$-{\left(bc\right)}^2-{\left(2bc\right)}^2-{\left(3bc\right)}^2-......-{(\frac{N}{bc}.bc)}^2$

......................................

$+{\left(abc\right)}^2+{\left(2abc\right)}^2+{\left(3abc\right)}^2+......+{(\frac{N}{abc}.abc)}^2$

.................................

Now, $a^2+{\left(2a\right)}^2+{\left(3a\right)}^2+......+{(\frac{N}{a}.a)}^2$

$=\frac{1}{6}{a}^2.\frac{N}{a}(\frac{N}{a}+1)(\frac{2N}{a}+1)$

$=\frac{N^3}{3a}+\frac{N^2}{2}+\frac{Na}{6};$

$\therefore$ The sum of the squares of all numbers less than $N$ and not prime to it is

$\frac{N^3}{3}\{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+....-\frac{1}{ab}-\frac{1}{ac}-.....+\frac{1}{abc}+...\}$

$+\frac{N^2}{2}\{m-\frac{m(m-1)}{1.2}+\frac{m(m-1)(m-2)}{\mathrm{1.2.3}}-.....\}$

$+\frac{N}{6}\{a+b+c+......-ab-ac-....+abc+...\}$

where m is no. of prime factors in N

Thus the co efficient of $\frac{N^2}{2}=1-{(1-1)}^m=1$

$\therefore$ The sum of the squares of all numbers less than $N$ and prime to it is obtained by subtracting the above expression form

$+\frac{N}{6}(N+1)(2N+1)$

or $\frac{N^3}{3}+\frac{N^2}{2}+\frac{N}{6}$

$\therefore$ The sum required is

$\frac{N^3}{3}\{1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}-....+\frac{1}{ab}+\frac{1}{ac}+.....-\frac{1}{abc}-...\}$

$+\frac{N}{6}\{1-a-b-c-......+ab+ac+....-abc-...\}$

$=\frac{N^3}{3}(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})....+\frac{N}{6}(1-a)(1-b)...$

To find the sum of the cubes;-

Let integers less than $N$ and prime to it be denoted by

$1,p,q,r,.....N-r,N-q,N-p,N-1$

If ${}^{\prime }{x}^{\prime }$ stands for any one of these integers, then

$\sum x^3=\sum {(N-x)}^3$ for each of these expressions denotes the sum of the same series of terms, only the order is $1$ is the reverse of that in the other,

Hence, $\sum x^3=\sum N^3-3\sum N^{2}x+3\sum x^{2}N-\sum x^3$

$2\sum x^3=N^3\Phi \left(N\right)-3N^2\sum x+3N\sum x^2$

$\Phi \left(N\right)\to$no. of terms

$\sum x=\frac{1}{2}N\Phi \left(N\right)$

$\sum x^2=\frac{N^2}{3}\Phi \left(N\right)+\frac{N}{6}(1-a)(1-b)(1-c)+......$

$2\sum x^3=\frac{N^3}{2}\Phi \left(N\right)+\frac{N^2}{2}(1-a)(1-b)(1-c)+....$

$\Phi \left(N\right)=N(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})....$