Question

Show that the sum of the fourth powers of all the numbers less than $N$ and prime to it is
$\frac{N^5}{5}(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})....+\frac{N^3}{3}(1-a)(1-b)(1-b)....$

$-\frac{N}{30}(1-a^3)(1-b^3)(1-c^3)....,a,b,c,....$ being the different prime factors of $N$.

Answer 1

$\Rightarrow$ We may show that the sum of the $r^{th}$ power of all integers less than $N$ amd prime to it is $S_N-a^r\frac{S_N}{a}-b^r\frac{S_N}{b}-......+{\left(ab\right)}^r\frac{S_N}{ab}+.....1;$

Where, $S_p=1^r+2^r+3^r+....+p^r$

$S_N=\frac{N^{r+1}}{r+1}+\frac{1}{2}{N}^r+B_1\frac{r}{2!}{N}^{r-1}-B_3\frac{r(r-1)(r-2)}{4!}{N}^{r-3}+....$

Hence, $x^r\frac{S_N}{6}=\frac{N^{r+1}}{r+1}\frac{1}{x}+\frac{1}{2}{N}^r+B_1\frac{r}{2!}{N}^{r-1}x-B_3\frac{r(r-1)(r-2)}{4!}{N}^{r-3}{x}^3+....$

$\therefore$ by substituting in $\left(1\right)$, we see that the sum of the $r^{th}$ powers of all integers less than $N$ and prime to it

$=\frac{N^{r+1}}{r+1}\{1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}-.....+\frac{1}{ab}+\}$

$+\frac{1}{2}{N}^r\{1-m+\frac{m(m-1)}{2!}-...\}$

$+B_1\frac{r}{2!}{N}^{r-1}\{1-a-b-c-....+ab+...\}$

$-B_3\frac{r(r-1)(r-2)}{4!}{N}^{r-3}\{1-a^3-b^3-c^3-....+a^3{b}^3+....\}+......;$

m is the no. of prime factors in $N$

Thus the co efficient of $N^r$ is $0$ and the sum required,

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

$+B_1\frac{r}{2!}{N}^{r-1}(1-a)(1-b)(1-c).....$

$-B_3\frac{r(r-1)(r-2)}{4!}{N}^{r-3}(1-a^3)(1-b^3)(1-c^3)+....$

If $r=2,B_1\frac{r}{2!}=\frac{1}{6}$

If $r=3,B_1\frac{r}{2!}=\frac{1}{4}$

If $r=4,B_1\frac{r}{2!}=\frac{1}{3},B_3\frac{r(r-1)(r-2)}{4!}=\frac{1}{30}$

Hence, by substitution we have

$\sum N^4=\frac{N^5}{5}(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})....$

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

$-\frac{N}{30}(1-a^3)(1-b^3)(1-c^3)+....$