Multiply Perfect Number of Order 6

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Theorem

The number defined as:

$n = 2^{36} \times 3^8 \times 5^5 \times 7^7 \times 11 \times 13^2 \times 19 \times 31^2$
$\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
$\times \ 7019 \times 112 \, 303 \times 898 \, 423 \times 616 \, 318 \, 177$

is multiply perfect of order $6$.


Proof

From Divisor Sum Function is Multiplicative, we may take each prime factor separately and form $\map {\sigma_1} n$ as the product of the divisor sum of each.


Each of the prime factors which occur with multiplicity $1$ will be treated first.

A prime factor $p$ contributes towards the combined $\sigma_1$ a factor $p + 1$.


Hence we have:

\(\ds \map {\sigma_1} {11}\) \(=\) \(\ds 12\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 3\)


\(\ds \map {\sigma_1} {19}\) \(=\) \(\ds 20\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 5\)


\(\ds \map {\sigma_1} {43}\) \(=\) \(\ds 44\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 11\)


\(\ds \map {\sigma_1} {61}\) \(=\) \(\ds 62\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2 \times 31\)


\(\ds \map {\sigma_1} {83}\) \(=\) \(\ds 84\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 3 \times 7\)


\(\ds \map {\sigma_1} {223}\) \(=\) \(\ds 224\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^5 \times 7\)


\(\ds \map {\sigma_1} {331}\) \(=\) \(\ds 332\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 83\)


\(\ds \map {\sigma_1} {379}\) \(=\) \(\ds 380\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 5 \times 19\)


\(\ds \map {\sigma_1} {601}\) \(=\) \(\ds 602\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2 \times 7 \times 43\)


\(\ds \map {\sigma_1} {757}\) \(=\) \(\ds 758\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2 \times 379\)


\(\ds \map {\sigma_1} {1201}\) \(=\) \(\ds 1202\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2 \times 601\)


\(\ds \map {\sigma_1} {7019}\) \(=\) \(\ds 7020\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^2 \times 3^3 \times 5 \times 13\)


\(\ds \map {\sigma_1} {112 \, 303}\) \(=\) \(\ds 112 \, 304\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^4 \times 7019\)


\(\ds \map {\sigma_1} {898 \, 423}\) \(=\) \(\ds 898 \, 424\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2^3 \times 112 \, 303\)


\(\ds \map {\sigma_1} {616 \, 318 \, 177}\) \(=\) \(\ds 616 \, 318 \, 178\) Divisor Sum of Prime Number
\(\ds \) \(=\) \(\ds 2 \times 7^3 \times 898 \, 423\)


The remaining factors are treated using Divisor Sum of Power of Prime:

$\map {\sigma_1} {p^k} = \dfrac {p^{k + 1} - 1} {p - 1}$


Thus:

\(\ds \map {\sigma_1} {2^{36} }\) \(=\) \(\ds 2 \times 2^{36} - 1\) Divisor Sum of Power of 2
\(\ds \) \(=\) \(\ds 137 \, 438 \, 953 \, 471\)
\(\ds \) \(=\) \(\ds 223 \times 616 \, 318 \, 177\)


\(\ds \map {\sigma_1} {3^8}\) \(=\) \(\ds \dfrac {3^9 - 1} {3 - 1}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds \dfrac {19 \, 683 - 1} 2\)
\(\ds \) \(=\) \(\ds 9841\)
\(\ds \) \(=\) \(\ds 13 \times 757\)


\(\ds \map {\sigma_1} {5^5}\) \(=\) \(\ds \dfrac {5^6 - 1} {5 - 1}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds \dfrac {15 \, 625 - 1} 4\)
\(\ds \) \(=\) \(\ds 3906\)
\(\ds \) \(=\) \(\ds 2 \times 3^2 \times 7 \times 31\)


\(\ds \map {\sigma_1} {7^7}\) \(=\) \(\ds \dfrac {7^8 - 1} {7 - 1}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds \dfrac {5 \, 764 \, 801 - 1} 6\)
\(\ds \) \(=\) \(\ds 960 \, 800\)
\(\ds \) \(=\) \(\ds 2^5 \times 5^2 \times 1201\)


\(\ds \map {\sigma_1} {13^2}\) \(=\) \(\ds \dfrac {13^3 - 1} {13 - 1}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds \dfrac {2197 - 1} {12}\)
\(\ds \) \(=\) \(\ds 183\)
\(\ds \) \(=\) \(\ds 3 \times 61\)


\(\ds \map {\sigma_1} {31^2}\) \(=\) \(\ds \dfrac {31^3 - 1} {31 - 1}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds \dfrac {29 \, 791 - 1} {30}\)
\(\ds \) \(=\) \(\ds 993\)
\(\ds \) \(=\) \(\ds 3 \times 331\)


Gathering up the prime factors, we have:

$\map {\sigma_1} n = 2^{37} \times 3^9 \times 5^5 \times 7^7 \times 11 \times 13^2 \times 19 \times 31^2$
$\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
$\times \ 7019 \times 112 \, 303 \times 898 \, 423 \times 616 \, 318 \, 177$


By inspection of the multiplicities of the prime factors of $n$ and $\map {\sigma_1} n$, it can be seen that they match for all except for $2$ and $3$.

It follows that $\map {\sigma_1} n = 2 \times 3 \times n = 6 n$.

Hence the result.

$\blacksquare$


Historical Note

Marin Mersenne, in a letter of $1643$, challenged Pierre de Fermat to find the ratio of:

$2^{36} \times 3^8 \times 5^5 \times 11 \times 13^2 \times 19 \times 31^2$
$\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
$\times \ 7019 \times 823 \, 543 \times 616 \, 318 \, 177 \times 100 \, 895 \, 598 \, 169$

to its aliquot sum.

Fermat replied that its ratio to the sum of all its divisors (including the number itself) was $6$.

He also pointed out that $100 \, 895 \, 598 \, 169 = 112 \, 303 \times 898 \, 423$, both of which divisors are prime.

Also note that $823 \, 543 = 7^7$, another point that Marin Mersenne glossed over, intentionally or inadvertently, in his initial challenge.

Both Mersenne's initial challenge and Fermat's factorisation of $100 \, 895 \, 598 \, 169$ were remarkable, considering the lack of computing machines in those days. To this day, nobody knows how they did it.


Sources

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