determining a prime factorization

Question

What is the prime factorizatíon of

$$N:= \sum_{n=1}^{10^8 -1} n^3 $$

writing the sum out, didn't help me.. therefore..any help is really appreciated!

Answer 1

In general, $\sum_{n=1}^k n^3 =\frac {k^2(k+1)^2}4$

In your case $k=10^8 -1$

The sum is $\frac {(10^8-1)^2(10^8)^2}4$

$10^8-1=99999999=3^2 \times 11 \times 73 \times 101 \times 137$

$\frac {(10^8)^2}4=\frac {10^{16}}4=\frac{2^{16} \times 5^{16}}{2^2}=2^{14} \times 5^{16}$

So $\frac {(10^8-1)^2(10^8)^2}4=2^{14} \times 3^4 \times 5^{16} \times 11^2 \times 73^2 \times 101^2 \times 137^2 $