# Challenge Math Challenge - March 2019

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#### DeathByKugelBlitz

Gold Member
Python also has "rstrip", which makes it a one-liner:
Code:
>>> len(str(math.factorial(1000)))-len(str(math.factorial(1000)).rstrip("0"))
249
This is what I was aiming for

#### Bosko

Gold Member
5.) Prove that starting with $\frac{1}{1}$ the following binary tree
$$\begin{array}{ccccc} &&\frac{a}{b}&&\\ &\swarrow &&\searrow \\ \frac{a}{a+b}&&&&\frac{a+b}{b} \end{array}$$
defines a counting of all positive rational numbers without repetition and all quotients canceled out.
1) Let prove that for any $\frac{a}{b}$ in the tree , a and b are positive natural numbers. ( The mathematical induction by depth n of the tree )
for 0 , $\frac{1}{1}$ , a=1 and b=1 are positive natural numbers
for n-1 , we assume that for all numbers in form $\frac{a_i }{b_i}$ on depth n-1 , $a_i$ and $b_i$ are positive and natural . ( depth = distance from the root )

for n , and any $\frac{a}{b}$ on depth n is either $a=a_i$ and $b=a_i + b_i$ or $a=a_i+b_i$ and $b=b_i$
In both cases a and b are positive natural numbers.

2) There is no non-root $\frac{c}{c}$ , in the tree (c is positive natural number as proven in (1.) ),
Let assume the opposite . There is non-root $\frac{c}{c}$ in the tree.
Then the "parent" $\frac{a}{b}$ of $\frac{c}{c}$ exists and either $\frac{c}{c}=\frac{a}{a+b}$ or $\frac{c}{c}=\frac{a+b}{b}$

if $\frac{c}{c}=\frac{a}{a+b}$
c=a , c=a+b
a=c , b=c-c
"parent" is $\frac{c}{0}$ -> 0 is not positive natural number -> contradiction

if $\frac{c}{c}=\frac{a+b}{b}$
c=a+b , c=b
a=c-c , b=c
"parent" is $\frac{0}{c}$ -> 0 is not positive natural number -> contradiction.

Consequence -> Only the "root" $\frac{1}{1}$ is equal 1
Notice -> Every "left child" is $\frac{a}{a+b}<1$ and every "right child" $\frac{a+b}{b}>1$ of any "parent" $\frac{a}{b}$

3) "... all quotients canceled out"
a) if c is divisor of a and a+b ( "left child" $\frac{a}{a+b}$ ) then c is divisor of a and b=a+b-a ("parent" $\frac{a}{b}$ )
b) if c is divisor of a+b and b ( "right child" $\frac{a+b}{b}$ ) then c is divisor of a=a+b-b and b ("parent" $\frac{a}{b}$ )
If c is divisor of $a_n$ and $b_n$ for some $\frac{a_n}{b_n}$ on the "depth" n of the tree.
Then c is divisor of $a_{n-1}$ and $b_{n-1}$ for the "parent" $\frac{a_{n-1}}{b_{n-1}}$ on the depth n-1 of the tree
... on the depth n-2 of the tree
...
Then c is divisor of $a_0=1$ and $b_0=1$ for the "root" $\frac{1}{1}$ on the depth 0 of the tree
Then c=1 . $\frac{a_n}{b_n}$ are already canceled out.

4) "... all positive rational numbers..." are in the tree
Let assume that there is a positive rational number $\frac{a_0}{b_0}$ not in the tree

If $a_0 < b_0$ then its "parent" $\frac{a_1}{b_1}=\frac{a_0}{b_0-a_0}$ is not in the tree also.
Because if the "parent" $\frac{a_1}{b_1}$ is in the tree then the "left child" $\frac{a_1}{a_1 + b_1}=\frac{a_0}{b_0-a_0+b_0}=\frac{a_0}{b_0}$ is in the tree
$a_0 = a_1$
$b_0 > b_1 = b_0-a_0$

If $a_0 > b_0$ then its "parent" $\frac{a_1}{b_1}=\frac{a_0-b_0}{b_0}$ is not in the tree also.
Because if the "parent" $\frac{a_1}{b_1}$ is in the tree then the "right child" $\frac{a_1+b_1}{b_1}=\frac{a_0-b_0+b_0}{b_0}=\frac{a_0}{b_0}$ is in the tree
$a_0 > a_1=a_0-b_0$
$b_0 = b_1$

Similarly we will get that the "parent" of the "parent" $\frac{a_2}{b_2}$ and $\frac{a_3}{b_3}$... are not in the tree
But in every step either ...
$a_i = a_{i+1}$ and $b_i> b_{i+1}$ or ...
$a_i > a_{i+1}$ and $b_i= b_{i+1}$
Means that $a_i + b_i > a_{i+1} + b_{i+1}$. and there is n so $a_n=b_n=1$. and $\frac{a_n}{b_n}=\frac{1}{1}$ is not in the tree .
but $\frac{a_n}{b_n}=\frac{1}{1}$ is the "root" of the tree.
Our assumption that $\frac{a_0}{b_0}$ is not in the tree is wrong

5) "... without repetition...." (as you assume / I will assume the opposite :-) )
There are two different "node"-s in the tree with the same numbers $\frac{a_0}{b_0}=\frac{c_0}{d_0}$
and its "parents" $\frac{a_1}{b_1}=\frac{c_1}{d_1}$ are equal and different nodes of the tree
and the "grand parents"...
...
and $\frac{1}{1}=\frac{a_n}{b_n}=\frac{c_n}{d_n}$ are equal and different nodes of the tree.
Our tree have two roots -> impossible . My assumption is wrong.

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#### fresh_42

Mentor
2018 Award
1) Let prove that for any $\frac{a}{b}$ in the tree , a and b are positive natural numbers. ( The mathematical induction by depth n of the tree )
for 0 , $\frac{1}{1}$ , a=1 and b=1 are positive natural numbers
for n-1 , we assume that for all numbers in form $\frac{a_i }{b_i}$ on depth n-1 , $a_i$ and $b_i$ are positive and natural . ( depth = distance from the root )

for n , and any $\frac{a}{b}$ on depth n is either $a=a_i$ and $b=a_i + b_i$ or $a=a_i+b_i$ and $b=b_i$
In both cases a and b are positive natural numbers.

2) There is no non-root $\frac{c}{c}$ , in the tree (c is positive natural number as proven in (1.) ),
Let assume the opposite . There is non-root $\frac{c}{c}$ in the tree.
Then the "parent" $\frac{a}{b}$ of $\frac{c}{c}$ exists and either $\frac{c}{c}=\frac{a}{a+b}$ or $\frac{c}{c}=\frac{a+b}{b}$

if $\frac{c}{c}=\frac{a}{a+b}$
c=a , c=a+b
a=c , b=c-c
"parent" is $\frac{c}{0}$ -> 0 is not positive natural number -> contradiction

if $\frac{c}{c}=\frac{a+b}{b}$
c=a+b , c=b
a=c-c , b=c
"parent" is $\frac{0}{c}$ -> 0 is not positive natural number -> contradiction.
No quite correct, as you cannot conclude that the quotients have equal numerators and denominators as long as you don't know that they are cancelled out, which for $c/c$ is obviously not the case.
Instead you have e.g. $c(a+b)=ca \Longrightarrow cb = 0$ which cannot be the case, as all numbers are positive.
Consequence -> Only the "root" $\frac{1}{1}$ is equal 1
Notice -> Every "left child" is $\frac{a}{a+b}<1$ and every "right child" $\frac{a+b}{b}>1$ of any "parent" $\frac{a}{b}$

3) "... all quotients canceled out"
a) if c is divisor of a and a+b ( "left child" $\frac{a}{a+b}$ ) then c is divisor of a and b=a+b-a ("parent" $\frac{a}{b}$ )
b) if c is divisor of a+b and b ( "right child" $\frac{a+b}{b}$ ) then c is divisor of a=a+b-b and b ("parent" $\frac{a}{b}$ )
If c is divisor of $a_n$ and $b_n$ for some $\frac{a_n}{b_n}$ on the "depth" n of the tree.
Then c is divisor of $a_{n-1}$ and $b_{n-1}$ for the "parent" $\frac{a_{n-1}}{b_{n-1}}$ on the depth n-1 of the tree
... on the depth n-2 of the tree
...
Then c is divisor of $a_0=1$ and $b_0=1$ for the "root" $\frac{1}{1}$ on the depth 0 of the tree
Then c=1 . $\frac{a_n}{b_n}$ are already canceled out.

4) "... all positive rational numbers..." are in the tree
Let assume that there is a positive rational number $\frac{a_0}{b_0}$ not in the tree

If $a_0 < b_0$ then its "parent" $\frac{a_1}{b_1}=\frac{a_0}{b_0-a_0}$ is not in the tree also.
Because if the "parent" $\frac{a_1}{b_1}$ is in the tree then the "left child" $\frac{a_1}{a_1 + b_1}=\frac{a_0}{b_0-a_0+b_0}=\frac{a_0}{b_0}$ is in the tree
$a_0 = a_1$
$b_0 > b_1 = b_0-a_0$

If $a_0 > b_0$ then its "parent" $\frac{a_1}{b_1}=\frac{a_0-b_0}{b_0}$ is not in the tree also.
Because if the "parent" $\frac{a_1}{b_1}$ is in the tree then the "right child" $\frac{a_1+b_1}{b_1}=\frac{a_0-b_0+b_0}{b_0}=\frac{a_0}{b_0}$ is in the tree
$a_0 > a_1=a_0-b_0$
$b_0 = b_1$

Similarly we will get that the "parent" of the "parent" $\frac{a_2}{b_2}$ and $\frac{a_3}{b_3}$... are not in the tree
But in every step either ...
$a_i = a_{i+1}$ and $b_i> b_{i+1}$ or ...
$a_i > a_{i+1}$ and $b_i= b_{i+1}$
Means that $a_i + b_i > a_{i+1} + b_{i+1}$. and there is n so $a_n=b_n=1$. and $\frac{a_n}{b_n}=\frac{1}{1}$ is not in the tree .
but $\frac{a_n}{b_n}=\frac{1}{1}$ is the "root" of the tree.
Our assumption that $\frac{a_0}{b_0}$ is not in the tree is wrong

5) "... without repetition...." (as you assume / I will assume the opposite :-) )
There are two different "node"-s in the tree with the same numbers $\frac{a_0}{b_0}=\frac{c_0}{d_0}$
and its "parents" $\frac{a_1}{b_1}=\frac{c_1}{d_1}$ are equal and different nodes of the tree
and the "grand parents"...
...
and $\frac{1}{1}=\frac{a_n}{b_n}=\frac{c_n}{d_n}$ are equal and different nodes of the tree.
Our tree have two roots -> impossible . My assumption is wrong.
Very nice.

For the sake of completion:
The recursive arguments can be shortened by the following "trick":

We define a norm of these elements by $N(p/q)=p+q\,.$ The parent quotient of $\frac{p}{q}\neq \frac{1}{1}$ is either $\frac{p}{q-p}$ or $\frac{p-q}{q}$. The norm of the child has strictly increased in both cases. Now we have a linear ordering of the levels and can argue by minimality instead of running all the way up.

#### Bosko

Gold Member
... We define a norm of these elements by $N(p/q)=p+q\,.$ The parent quotient of $\frac{p}{q}\neq \frac{1}{1}$ is either $\frac{p}{q-p}$ or $\frac{p-q}{q}$. The norm of the child has strictly increased in both cases. Now we have a linear ordering of the levels and can argue by minimality instead of running all the way up.
Nice idea . Thanks

For me is very interesting that all pairs a/b in the tree are mutually prime ("co-prime" GDC(a,b)=1 )
And of course all mutually prime positive pairs (a, b) are somewhere in the tree as a/b.
We can make the algorithm that generate two random a, b and we know that they are mutually prime without checking.
Code:
a=1
b=1

repeat ...
r= random () # random number from the segment[0,1]
if r<0.5 then
b=a+b
else
a=a+b
Maybe this can be used in the RSA or in the other type of the encryption , or for a something else ...

#### fbs7

My attempt at High School 4; notice I had to spy on Euler's proof for directions, so I'm not sure if this an acceptable answer or not:

=====================================================================
I) if (2^n-1) is prime, then z=2^(n-1)*(2^n-1) is perfect:

x1 = 2^0*(2^1-1) = 2
x2 = 2^1*(2^2-1) = 2*3 = 12
x3 = 2^2*(2^3-1) = 4*7 = 28
...
xk = 2^(n-1)*(2^n-1)

factors are

1,2,2^2,...,2^(n-1),
(2^n-1),2*(2^n-1),(2^2)*(2^n-1),...,2^(n-2)*(2^n-1)

sum of the factors (less z) is

S=1+2^1+2^2+...+2^(n-1)+
(2^n-1)*(2^0+2^1+...+2^(n-2))

the sum of a geometric series is a1*(r^n-1)/(r-1); so

S=1*(2^n-1)/(2-1) + (2^n-1)*( 1*(2^(n-1)-1)/(2-1) )
= (2^n-1) + (2^n-1)*(2^(n-1)-1)
= (2^n-1)*(2^(n-1)-1+1)
= (2^n-1)*2^(n-1)=z

so the sum of factors (less z) is equal to z, so z is perfect.
============================================================================
II) If X is even and X is perfect, then X = (2^n-1)*2^(n-1)

I had to spy on Euler's proof for this guy; this is my version of Euler's proof, assuming no previous knowledge.

divisors of x = {1,x1,x2,...} = {xi}
divisors of y = {1,y1,y2,...} = {yj}

define s(x) as the sum of all divisors of x (including x), ie
s(x) = sum( xi )
s(y) = sum( yj )

if z = x.y
then every divisor of z has the form xi*yj, therefore

s(z) = (1*1+1*y1+1*y2+...+x1*1+x1*y1+x1*y2+...+x2*1+x2*y1+x2*y2+...
= 1*(1+y1+y2+...)+x1*(1+y1+y2+...)+x2*(1+y1+y2+...)+...
= 1*s(y)+x1*s(y)+x2*s(y)+...
= (1+x1+x2+...)*s(y)
= s(x)*s(y)

therefore s(x.y) = s(x)*s(y)

So, if z is even and perfect, then z is a multiple of 2, ie for some k there is an odd x:

z = 2^(k-1)*x, x is odd
s(z) = s( 2^(k-1) * x ) = s(2^(k-1))*s(x)
for k=1, we have s(2^(k-1)) = s(1) = 1
for k=2, we have s(2^(k-1)) = s(2) = 1+2
for k=3, we have s(2^(k-1)) = s(3) = 1+2+4
...
for k=k, we have s(2^(k-1)) = 1+2^1+...+2^(k-1)

that's a geometric series with a1=1, r=2, n=k; ie

s(2^(k-1)) = 1*(2^k-1)/(2-1) = 2^k-1
s(z) = (2^k-1)*s(x)

as z is perfect, s(z) = 2*z, so

s(z) = 2*z = 2*2^(k-1)*x = (2^k-1)*s(x)
(2*2^(k-1))*x = (2^k-1)*s(x)
(2^k)*x = (2^k-1)*s(x)
(2^k-1)*x + x = (2^k-1)*s(x)
s(x) = x + x/(2^k-1)

s(x) is integer and x is integer, therefore x/(2^k-1) is integer, therefore (2^k-1) is a divisor of x.

It remains to prove that x has no other divisors except 1. For some y,
x = (2^k-1)*y
s(x) = x + (2^k-1)*y/(2^k-1) = x + y

now, the divisors of x will in general be {1,a,b,c,...,x }, therefore
s(x) = 1+a+b+c+...+x = x+y

that's impossible to satisfy for any a,b,c, therefore a, b, c do not exist, and x only has 2 divisors, so:
s(x) = 1 + x = x + y

then y=1, and x = (2^k-1) and has no other divisors other than 1 and itself, therefore it is a prime. That way
z = (2^k-1)*(2^k-1), where 2^k-1 is prime

#### fbs7

Hmm... there must be another way of doing 4, without cheating with Euler... hmm...

#### fresh_42

Mentor
2018 Award
Hmm... there must be another way of doing 4, without cheating with Euler... hmm...
You should use the hint of the question before, they are closely related.

Could you please transform your text into LaTeX format? It is hard to read and one never knows for sure whether you meant $2^{n-1}$ or $2^n-1$.
Indices are written x_1 and not x1, formulas with a double sharp $2^{n-1} \text{ or } 2^n-1$ and not 2^n-1.

See https://www.physicsforums.com/help/latexhelp/, it's really not difficult - at least much easier than to decode your linear text.

#### fbs7

Hmm...sorry... trying with the latex thingie with my cheater's proof:

I) if $(2^n-1)$ is prime, then $z=2^{(n-1)}*(2^n-1)$ is perfect:

the factors of z (less z) are:

$1,2,2^2,...,2^{(n-1)},$
$(2^n-1),2*(2^n-1),(2^2)*(2^n-1),...,2^{(n-2)}*(2^n-1)$

sum of the factors (less z) is

$S=1+2^1+2^2+...+2^{(n-1)}+$
$(2^n-1)*(2^0+2^1+...+2^{(n-2)})$

the sum of a geometric series is $a_1*(r^n-1)/(r-1)$; so

$S=1*(2^n-1)/(2-1) + (2^n-1)*( 1*(2^{(n-1)}-1)/(2-1) )$
$= (2^n-1) + (2^n-1)*(2^{(n-1)}-1)$
$= (2^n-1)*(2^{(n-1)}-1+1)$
$= (2^n-1)*2^{(n-1)}=z$

so the sum of factors (less z) is equal to z, so z is perfect.
============================================================================
II) If X is even and X is perfect, then $X = (2^n-1)*2^{(n-1)}$

I had to spy on Euler's proof for this guy; this is my version of Euler's proof, assuming no previous knowledge.

divisors of $x =${$1,x_1,x_2,...,x$} = {$x_i$}
divisors of $y =${$1,y_1,y_2,...,y$} = {$y_j$}

define s(x) as the sum of all divisors of x (including x), ie
$s(x) = sum( x_i )$
$s(y) = sum( y_j )$

if $z = x*y$
then every divisor of z has the form $x_i*y_j$, therefore

$s(z) = (1*1+1*y_1+1*y_2+...+x_1*1+x_1*y_1+x_1*y_2+...+x_2*1+x_2*y_1+x_2*y_2+...$
$= 1*(1+y_1+y_2+...)+x_1*(1+y_1+y_2+...)+x_2*(1+y_1+y_2+...)+...$
$= 1*s(y)+x_1*s(y)+x_2*s(y)+...$
$= (1+x_1+x_2+...)*s(y)$
$= s(x)*s(y)$

therefore $s(x*y) = s(x)*s(y)$

So, if z is even and perfect, then z is a multiple of 2, ie for some k there is an odd x:

$z = 2^{(k-1)}*x$, x is odd
$s(z) = s( 2^{(k-1)} * x ) = s(2^(k-1))*s(x)$
for k=1, we have $s(2^{(k-1)}) = s(1) = 1$
for k=2, we have $s(2^{(k-1)}) = s(2) = 1+2$
for k=3, we have $s(2^{(k-1)}) = s(4) = 1+2+4$
...
for k=k, we have $s(2^{(k-1)}) = 1+2^1+...+2^{(k-1)}$

that's the sum of a geometric series with a1=1, r=2, n=k; ie

$s(2^{(k-1)}) = 1*(2^k-1)/(2-1) = 2^k-1$
$s(z) = (2^k-1)*s(x)$

as z is perfect, s(z) = 2*z, so

$s(z) = 2*z = 2*2^{(k-1)}*x = (2^k-1)*s(x)$
$(2*2^{(k-1)})*x = (2^k-1)*s(x)$
$(2^k)*x = (2^k-1)*s(x)$
$(2^k-1)*x + x = (2^k-1)*s(x)$
$s(x) = x + x/(2^k-1)$

s(x) is integer and x is integer, therefore $x/(2^k-1)$ is integer, therefore $(2^k-1)$ is a divisor of x.

It remains to prove that x has no other divisors except 1. For some y,
$x = (2^k-1)*y$
$s(x) = x + (2^k-1)*y/(2^k-1) = x + y$

now, the divisors of x will in general be {1,a,b,c,...,x }, therefore
$s(x) = 1+a+b+c+...+x = x+y$

that's impossible to satisfy for any a,b,c, therefore a, b, c do not exist, and x only has 2 divisors, so:
$s(x) = 1 + x = x + y$

then $y=1$, and $x = (2^k-1)$ and has no other divisors other than 1 and itself, therefore it is a prime. That way
$z = (2^k-1)*(2^{k-1})$, where $2^{k-1}$ is prime

I so wish there's a way to prove this by induction; this is kinda screaming induction... Induction would be so awesome!

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#### fresh_42

Mentor
2018 Award
Looks o.k. in general. Just a few remarks. I think there are some further steps needed.
Hmm...sorry... trying with the latex thingie with my cheater's proof:
I) if $(2^n-1)$ is prime, then $z=2^{(n-1)}*(2^n-1)$ is perfect:

the factors of z (less z) are:

$1,2,2^2,...,2^{(n-1)},$
$(2^n-1),2*(2^n-1),(2^2)*(2^n-1),...,2^{(n-2)}*(2^n-1)$

sum of the factors (less z) is

$S=1+2^1+2^2+...+2^{(n-1)}+$
$(2^n-1)*(2^0+2^1+...+2^{(n-2)})$

the sum of a geometric series is $a_1*(r^n-1)/(r-1)$; so

$S=1*(2^n-1)/(2-1) + (2^n-1)*( 1*(2^{(n-1)}-1)/(2-1) )$
$= (2^n-1) + (2^n-1)*(2^{(n-1)}-1)$
$= (2^n-1)*(2^{(n-1)}-1+1)$
$= (2^n-1)*2^{(n-1)}=z$

so the sum of factors (less z) is equal to z, so z is perfect.
============================================================================
II) If X is even and X is perfect, then $X = (2^n-1)*2^{(n-1)}$

I had to spy on Euler's proof for this guy; this is my version of Euler's proof, assuming no previous knowledge.

divisors of $x =${$1,x_1,x_2,...,x$} = {$x_i$}
divisors of $y =${$1,y_1,y_2,...,y$} = {$y_j$}

define s(x) as the sum of all divisors of x (including x), ie
$s(x) = sum( x_i )$
$s(y) = sum( y_j )$

if $z = x*y$
then every divisor of z has the form $x_i*y_j$, therefore
It is clear that all $x_i \cdot y_j$ divide $x \cdot y$. But why does every divisor have this form?
Given $d\,|\,(xy)$, show that $d=x_iy_j$ for some pair $(i,j)\,?$

The problem is the following: If $d|(xy)$ then $d=x_{i_1}\ldots x_{i_k}y_{j_1}\ldots y_{j_l}$, i.e. the divisors of $d$ are somehow distributed over the divisors of $x$ and $y$. But we cannot conclude that $x_{i_1}\ldots x_{i_k}$ is again a divisor of $x$, e.g. $2|12$ and $4|12$ but $8\nmid 12\,.$ So how does $d$ have the desired form?

In other words: $s(2)\cdot s(6) = 3 \cdot 12 = 36 \neq 28 = s(12)$. So $s(xy)\neq s(x)s(y)$ in general. However, with an additional assumption it is true, and you only need it in the restricted version. Which is the additional assumption and can you prove it in that case?
$s(z) = (1*1+1*y_1+1*y_2+...+x_1*1+x_1*y_1+x_1*y_2+...+x_2*1+x_2*y_1+x_2*y_2+...$
$= 1*(1+y_1+y_2+...)+x_1*(1+y_1+y_2+...)+x_2*(1+y_1+y_2+...)+...$
$= 1*s(y)+x_1*s(y)+x_2*s(y)+...$
$= (1+x_1+x_2+...)*s(y)$
$= s(x)*s(y)$

therefore $s(x*y) = s(x)*s(y)$

So, if z is even and perfect, then z is a multiple of 2, ie for some k there is an odd x:

$z = 2^{(k-1)}*x$, x is odd
$s(z) = s( 2^{(k-1)} * x ) = s(2^(k-1))*s(x)$
for k=1, we have $s(2^{(k-1)}) = s(1) = 1$
for k=2, we have $s(2^{(k-1)}) = s(2) = 1+2$
for k=3, we have $s(2^{(k-1)}) = s(4) = 1+2+4$
...
for k=k, we have $s(2^{(k-1)}) = 1+2^1+...+2^{(k-1)}$

that's the sum of a geometric series with a1=1, r=2, n=k; ie

$s(2^{(k-1)}) = 1*(2^k-1)/(2-1) = 2^k-1$
$s(z) = (2^k-1)*s(x)$

as z is perfect, s(z) = 2*z, so

$s(z) = 2*z = 2*2^{(k-1)}*x = (2^k-1)*s(x)$
$(2*2^{(k-1)})*x = (2^k-1)*s(x)$
$(2^k)*x = (2^k-1)*s(x)$
$(2^k-1)*x + x = (2^k-1)*s(x)$
$s(x) = x + x/(2^k-1)$

s(x) is integer and x is integer, therefore $x/(2^k-1)$ is integer, therefore $(2^k-1)$ is a divisor of x.

It remains to prove that x has no other divisors except 1. For some y,
$x = (2^k-1)*y$
$s(x) = x + (2^k-1)*y/(2^k-1) = x + y$

now, the divisors of x will in general be {1,a,b,c,...,x }, therefore
$s(x) = 1+a+b+c+...+x = x+y$

that's impossible to satisfy for any a,b,c, therefore a, b, c do not exist, and x only has 2 divisors, so:
$s(x) = 1 + x = x + y$
Why is it impossible? What happens if $a>1\,$?
then $y=1$, and $x = (2^k-1)$ and has no other divisors other than 1 and itself, therefore it is a prime. That way
$z = (2^k-1)*(2^{k-1})$, where $2^{k-1}$ is prime

I so wish there's a way to prove this by induction; this is kinda screaming induction... Induction would be so awesome!
The proof would have been a bit shorter with a closed formula for $s(z)$.

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#### fbs7

Looks o.k. in general. Just a few remarks. I think there are some further steps needed.

It is clear that all $x_i \cdot y_j$ divide $x \cdot y$. But why does every divisor have this form?
Given $d\,|\,(xy)$, show that $d=x_iy_j$ for some pair $(i,j)\,?$

The problem is the following: If $d|(xy)$ then $d=x_{i_1}\ldots x_{i_k}y_{j_1}\ldots y_{j_l}$, i.e. the divisors of $d$ are somehow distributed over the divisors of $x$ and $y$. But we cannot conclude that $x_{i_1}\ldots x_{i_k}$ is again a divisor of $x$, e.g. $2|12$ and $4|12$ but $8\nmid 12\,.$ So how does $d$ have the desired form?

In other words: $s(2)\cdot s(6) = 3 \cdot 12 = 36 \neq 28 = s(12)$. So $s(xy)\neq s(x)s(y)$ in general. However, with an additional assumption it is true, and you only need it in the restricted version. Which is the additional assumption and can you prove it in that case?
Hmmm... s(2)=1+2=3, s(3)=1+3=4, s(2*3)=s(6)=1+2+3+6=12=s(2)*s(3)... in the same way s(2)=1+2=3, s(4)=1+2+4=7, s(2*4)=s(8)=1+2+4+8=15... holy choo-choo, that's different than s(2)*s(4)!!!! I forgot something!!! Thanks Lord that they don't let high-schoolers design homes, otherwise they would forget something like the doors and everybody would have to enter through the roof!

I suspect s(a*b)=s(a)*s(b) only if a and b have no common divisors other than 1 (no common factors). How to prove... for s(2) and s(4), the factors are {1,2} and {1,2,4}, therefore the $x_i*y_j$ forms would be {1*1, 1*2, 1*4, 2*1, 2*2, 2*4} = {1,2,4,2,4,8}... so the 2 and 4 are counted twice... that's because $x_i1*y_j1$ = $x_i2*y_i2$ for some i1 <> i2. If we separate these double-counted forms, then s(2)*(4)-(2+4)=(1+2+4+8)=15=s(8), so s(2*4) = s(2)*s(4) - M. So I suspect in general s(a*b) = s(a)*s(b) - M.

So, two troubles: how to prove M=0 if a and b have no common factors, and how to prove that if d|(x*y) then $d=x_i*y_j$. Hmm... let's try this without cheating (I cheated too much already by spying on Euler's!).

Defining $d|z$ is true if d is a divisor of z.

I) if d|(x*y) then d is a product of a factor of x and a factor of y; this can be proved by logical arguments (not sure how to prove it an algebraically correct way), as:

expressing z=x*y,d,x,y as the product of primes (notice these are different than $x_i, y_i$ and $z_i$ from the previous post!!: $z=2^{z_2}*3^{z_3}*5^{z_5}*..., d=2^{d_2}*3^{d_3}*5^{d_5}*..., x=2^{x_2}*3^{x_3}*5^{x_5}, y=2^{y_2}*3^{y_3}*5^{y_5}$, then for every prime factor p, $z_p=x_p+y_p$

as $d|z$, $d_p<=z_p$, ie $d_p<=x_p+y_p$; define:

$dx_p = min( x_p, d_p )$, what makes $p^{dx_p}|x$
$dy_p = d_p - dx_p$, what makes $p^{dy_p}|y$ [Note]

[Note: $d_p <= x_p+y_p$, if $dx_p=d_p$ then $dy_p=0$ ie $p^{dy_p}|y$; otherwise if $dx_p=x_p$ that means $d_p - dx_p <= x_p+y_p - x_p = y_p$, or $dy_p - dx_p <= y_p$ therefore $p^{dy_p}|y$]

then name

$u=2^{dx_2}*3^{dx_3}*...$, that makes $u|d$ and $u|x$
$v=2^{dy_2}*3^{dy_3}*...$, that means $v|d$ and $v|y$

we get d=u*v for the reason that $dx_p+dy_p=d_p$

so, if $d|(x*y)$, then there exists two numbers u and v, such that $d=u*v$ and $u|x$ and $v|z$. This holds true even if they have common factors.

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#### fresh_42

Mentor
2018 Award
Hmmm... s(2)=1+2=3, s(3)=1+3=4, s(2*3)=s(6)=1+2+3+6=12=s(2)*s(3)... in the same way s(2)=1+2=3, s(4)=1+2+4=7, s(2*4)=s(8)=1+2+4+8=15... holy choo-choo, that's different than s(2)*s(4)!!!! I forgot something!!! Thanks Lord that they don't let high-schoolers design homes, otherwise they would forget something like the doors and everybody would have to enter through the roof!

I suspect s(a*b)=s(a)*s(b) only if a and b have no common divisors other than 1 (no common factors).
That is a good suspicion! Such numbers are called coprime.
How to prove... for s(2) and s(4), the factors are {1,2} and {1,2,4}, therefore the $x_i*y_j$ forms would be {1*1, 1*2, 1*4, 2*1, 2*2, 2*4} = {1,2,4,2,4,8}... so the 2 and 4 are counted twice... that's because $x_i1*y_j1$ = $x_i2*y_i2$ for some i1 <> i2. If we separate these double-counted forms, then s(2)*(4)-(2+4)=(1+2+4+8)=15=s(8), so s(2*4) = s(2)*s(4) - M. So I suspect in general s(a*b) = s(a)*s(b) - M.

So, two troubles: how to prove M=0 if a and b have no common factors, and how to prove that if d|(x*y) then $d=x_i*y_j$. Hmm... let's try this without cheating (I cheated too much already by spying on Euler's!).

Defining $d|z$ is true if d is a divisor of z.

I) if d|(x*y) then d is a product of a factor of x and a factor of y; this can be proved by logical arguments (not sure how to prove it an algebraically correct way), as:

expressing z=x*y,d,x,y as the product of primes (notice these are different than $x_i, y_i$ and $z_i$ from the previous post!!: $z=2^{z_2}*3^{z_3}*5^{z_5}*..., d=2^{d_2}*3^{d_3}*5^{d_5}*..., x=2^{x_2}*3^{x_3}*5^{x_5}, y=2^{y_2}*3^{y_3}*5^{y_5}$, then for every prime factor p, $z_p=x_p+y_p$

as $d|z$, $d_p<=z_p$, ie $d_p<=x_p+y_p$; define:

$dx_p = min( x_p, d_p )$, what makes $p^{dx_p}|x$
$dy_p = d_p - dx_p$, what makes $p^{dy_p}|y$ [Note]

[Note: $d_p <= x_p+y_p$, if $dx_p=d_p$ then $dy_p=0$ ie $p^{dy_p}|y$; otherwise if $dx_p=x_p$ that means $d_p - dx_p <= x_p+y_p - x_p = y_p$, or $dy_p - dx_p <= y_p$ therefore $p^{dy_p}|y$]

then name

$u=2^{dx_2}*3^{dx_3}*...$, that makes $u|d$ and $u|x$
$v=2^{dy_2}*3^{dy_3}*...$, that means $v|d$ and $v|y$

we get d=u*v for the reason that $dx_p+dy_p=d_p$

so, if $d|(x*y)$, then there exists two numbers u and v, such that $d=u*v$ and $u|x$ and $v|z$. This holds true even if they have common factors.
Your proof is a bit dizzy for my taste. It is better to work with the definition of primes and the fundamental theorem of arithmetics, rather than the linear order.

The fundamental theorem of arithmetic says, that any integer can be written as a product of primes (and of course any number of factors $\pm 1$ which we do not count here).

A prime $p$ is a number, such if $p$ divides a product $a\cdot b$, then it divides necessarily one of the factors $a$ or $b$.
(Again, we rule out the units: $p\neq \pm 1$ and of course, $p$ can still divide both, $a$ and $b$, but it has to divide at least one of them.)

Now let us write $d = p_1^{r_1}\cdot \ldots \cdot p_m^{r_m}$. Then $p_1\,|\,d\,|\,(x\cdot y)$ and thus e.g. $p_1\,|\, x$. Now we can cancel $p_1$ and proceed with the next prime. At the end they are all either divisors of $x$ or of $y$. Since $x$ and $y$ are coprime, they do not share a prime and we have partitioned all divisors of $p$, either in the $x-$bucket or the $y-$bucket and most of all, did not count the same factor twice.

Hence you are allowed to apply the formula to $z=2^{k-1}\cdot x\, , \,x \text{ odd }$. The condition $x$ odd makes the factors coprime and the formula holds.

Can you explain the last statement, that $s(x) = 1+a+b+c+...+x = x+y$ implies $a =b=c=\ldots =0$, i.e. that $a>1$ is not possible?

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#### fbs7

II) How to prove that if x and y have no common divisors, then M=0 in s(x*y)=s(x)*s(y) - M

consider [$d_2,d_3,d_5,d_7,...$] as the factorization of d in primes: d=$2^{d_2}*3^{d_3}*...$; then

x=[$x_2,x_3,...$]
y=[$y_2,y_3,...$]
z=x*y=[$z_2,z_3,...$], $z_p=x_p+y_p$
d=[$d_2,d_3,...$], $d_p<=z_p$

each divisor u of x is a set of numbers [$u_2,u_3,...$] such that $u_p<=x_p$; same for each divisor v of y. When x and y have common divisors, u=v for some u and v.

The multiplication of each divisor u and v is a number w with prime factors $w_p=u_p+v_p <= x_p + y_p$; if x and y have no common divisors, then for each p $w_p = x_p$ or $x_p = y_p$; therefore each $d|z$ is a product of a unique pair (u,v) where $u|x$ and $v|y$.

Conversely, if x and y have common divisors, then there can be multiple pairs u,v such that d|z, d=u*v, u|x, v|y. Call

N(d,x,y) = number of pairs (u,v) such that d=u*v, u|x, v|y

s(x)*s(y) = sum every pair u*v such that u|x and v|y
s(x)*s(y) = sum( d * N(d,x,y) ) for every unique d such that d|(x*y)
s(x)*s(y) = sum(d) + sum(d)*(N(d,x,y)-1) for every d such that d|(x*y)

s(x)*s(y) = s(x*y) + sum( d*(N(d,x,y)-1 ) for every d such that d|(x*y)

if (x,y) have no common divisors, then for every d|(x*y), N(d,x,y) = 1, therefore M=0 and s(x)*s(y)=s(x*y)

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#### fbs7

Your proof is a bit dizzy for my taste. It is better to work with the definition of primes and the fundamental theorem of arithmetics, rather than the linear order.
I see, thank you!

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#### fbs7

Can you explain the last statement, that $s(x) = 1+a+b+c+...+x = x+y$ implies $a =b=c=\ldots =0$, i.e. that $a>1$ is not possible?
My reasoning there is 1 and x are divisors of x, therefore s(x) = 1+x+...other divisors...

We know that s(x) = x+y, that is, both x and y are divisors of x. But 1 is always a divisor of x, so

s(x) = 1+x+(y-1)

Assume y>1, then that makes y-1 also a divisor of x, therefore the divisors of x will be 1,x,y,y-1, so

s(x) = 1+x+y+y-1 = x + 2*y

but s(x) = x+y, so that's impossible with y>0, so the assumption is wrong and y<=1 and therefore y=1 and there are no other divisors a,b,c... of x.

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#### fresh_42

Mentor
2018 Award
II) How to prove that if x and y have no common divisors, then M=0 in s(x*y)=s(x)*s(y) - M

consider [$d_2,d_3,d_5,d_7,...$] as the factorization of d in primes: d=$2^{d_2}*3^{d_3}*...$; then

x=[$x_2,x_3,...$]
y=[$y_2,y_3,...$]
z=x*y=[$z_2,z_3,...$], $z_p=x_p+y_p$
d=[$d_2,d_3,...$], $d_p<=z_p$

each divisor u of x is a set of numbers [$u_2,u_3,...$] such that $u_p<=x_p$; same for each divisor v of y. When x and y have common divisors, u=v for some u and v.
....
This is very difficult to understand. What is $d$? You only have $s(xy)-s(x)s(y)=M$ and that $x$ and $y$ are coprime. That's where you start. Everything else has to be explained or defined. And then come $u$ and $v$ out of the blue. And how come the pairing $z_p=x_p+y_p\,?$
I'm completely confused.

My reasoning there is 1 and x are divisors of x, therefore s(x) = 1+x+...other divisors...

We know that s(x) = x+y, that is, both x and y are divisors of x. ....
Why that again? Sure, $x$ is a divisor of $x$, but why does $y$ have to be? E.g. $s(12)=28=12+16$ and $16$ does definitely not divide $12$.

Let me resume where we are:
The first part of your proof in post #68 was absolutely correct: If $2^n-1$ is prime, then $2^{n-1}\cdot (2^n-1)$ is perfect.

For the second part you have chosen a perfect number $z=2^{k-1}\cdot x$ with $k>1$ and $x$ odd.
What we now need is $$2z=2^k\cdot x = s(z)=s(2^{k-1}\cdot x) \stackrel{(1)}{=} s(2^{k-1}) \cdot s(x) = (2^k-1)\cdot s(x)$$ i.e. under the assumption that $(1)$ is correct, we have with $x=(2^k-1)\cdot y$ and your correct argument about integer values here
$$x+y \leq s(x)= 2^k\cdot y = (2^k-1)\cdot y + y =x+y$$
But now you have on the left the same as on the right which makes the inequality an equation, i.e. $s(x)=x+y$.

Let me put what you wrote in my own words, just to make it clear for myself. (I can better think if I write it and I have to keep in mind, that $s(a)=a+b$ does not mean per se that $b$ divides $a$, see my example above.)

We already know that $y$ divides $x$ per construction. So if $y \notin \{\,1,x\,\}$ then $s(x) \geq 1+x+y > x+y = s(x)$ which cannot be. $y=x$ is also impossible, since $k>1$ by assumption. Thus $y=1$ is the only remaining possibility, i.e.
... then $y=1$, and $x= 2^k-1$ and has no other divisors other than $1$ and itself, therefore it is a prime. That way $z = (2^k-1)\cdot 2^{k-1}$, where $2^k-1$ is prime. [edit: typo corrected]
So only $(1)$ is left to show:
Let $a$ be a power of $2$ and $b$ an odd number, then $s(a\cdot b) = s(a)\cdot s(b)$.

I think in this special case, your counting argument from post #68 will work, which might be the easiest way to show it. Another way is to solve the hint in the previous problem #3 of the High School section.

Functions which satisfy $s(xy)=s(x)\cdot s(y)$ for coprime $x,y$ are called multiplicative functions in number theory, and here we use that the sum of all divisors is a multiplicative function.

Numbers $2^k-1$ are called Mersenne numbers and primes $2^k-1$ Mersenne primes, in which case $k$ has to be prime, too. It is unclear (but suspected), whether there are infinitely many Mersenne primes. The highest known number is currently $2^{82,589,933}-1$ with $24,862,048$ digits. It is also unclear whether there are infinitely many perfect numbers.

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#### fbs7

This is very difficult to understand. What is ddd? You only have s(xy)−s(x)s(y)=M and that x and y are coprime. That's where you start. Everything else has to be explained or defined. And then come uuu and vvv out of the blue. And how come the pairing zp=xp+yp?zp=xp+yp?z_p=x_p+y_p\,?
I'm completely confused.
Sorry, I let mind to wander around in that post, as I wanted to get a result also for the case when x and y are not coprimes (it was rather easy to get it for when they are coprimes, so I wanted to go a bit beyond). Let me try to organize it. That factorization in prime factors was really superfluous.

$d|(x*y)$, d is a divisor of x*y
$u|x$, $v|y$, u is a divisor of x, v is a divisor of y

from the previous results we know that every
$d=u*v$ for some $u|x$, $v|y$

in general multiple $u,v$ satisfy $d=u*v$, so define
N(d,x,y) = number of $u,v$ pairs that satisfy $d=u*v, u|x, v|y$

$s(x) = \sum u$ for all $u|x$
$s(y) = \sum v$ for all $v|y$

$s(x)*s(y) = \sum \sum u*v$ for all $u|x, v|y$
$s(x)*s(y) = \sum N(d,x,y)*d$ for all distinct d such that $d=u*v, u|x, v|y$
$s(x)*s(y) = \sum d + \sum (N(d,x,y)-1)*d$ for all distinct d such that $d=u*v, u|x, v|y$

call
$S = \sum d$ for all distinct d such that $d=u*v, u|x, v|y$

as the sum occurs over all possible u,v pairs, that will include all possible d such that $d|(x*y)$
$S= \sum d$ for all distinct d such that $d|(x*y)$

by definition of s(),
$S = s(x*y)$
$s(x)*s(y) = s(x*y) - \sum (N(d,x,y)-1)*d$ for all distinct d such that $d=u*v, u|x, v|y$
$s(x)*s(y) = s(x*y) - M$
$M = \sum (N(d,x,y)-1)*d$

if x and y are co-primes then either $d=u*1$ or $d=1*v$, therefore N(d,x,y)=1 and M=0.

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#### fbs7

Bit of checking my equation for M... x=6, y=12...

factors of x=(1,2,3,6) => s(x) = 12
factors of y=(1,2,4,6,12) => s(y) = 25
factors of x*y=72=(1,2,3,4,6,8,12,18,24,36,72) => s(x*y) = 186

calculating the N(...); all of them are 1 except:
N(2,6,12)=2 from 1*2 and 2*1
N(4,6,12)=2 from 1*4 and 2*2
N(6,6,12)=3 from 1*6, 3*2 and 6*1
N(12,6,12)=4 from 1*12, 2*6, 3*4 and 6*1
N(24,6,12)=2 from 2*12 and 6*4
N(36,6,12)=2 from 3*12 and 6*6

so M=1*2+1*4+2*6+3*12+1*24+1*36=2+4+12+36+24+36=114
s(x)*s(y)-M = 12*25 - 114 = 300 - 114 = 186 = s(x*y)

Hoorye! I've found an equation!! Rather useless, but that's good work for a Sunday afternoon!

#### fresh_42

Mentor
2018 Award
You need to quantify your variables: for any / for all or there exist. Otherwise it is not clear what you mean. E.g.
$d|(x*y)$, $d$ is a divisor of $x*y$
$u|x$, $v|y$, $u$ is a divisor of $x$, $v$ is a divisor of $y$

from the previous results we know that every
$d=u*v$ for some $u|x$, $v|y$
Every divisor $d$ of the form $d=uv$ ... what? If $u,v$ are as given, then $d:=uv$ divides the product? All divisors are of this form? We only know this in case $x$ and $y$ are coprime, but you just said, that you want to consider the general case.

The art of writing a proof is the following: explain it step by step to someone who doesn't know where your thoughts currently are, like you would explain it to a stranger. Both is actually the case if you write something here. If you practice this you will find another advantage: the more detailed you explain something, always asking yourself why before the other one can, then you often learn more than you would learn by reading a book. The necessity to convince someone and thereby always convince yourself just a moment earlier can give you more insights than you might think. I once had a professor who I had the following dialogue with:
Me: I've seen you announced a lecture on measure theory next semester. I didn't know you were an expert.
Him: I'm not. I want to learn it.

For the task above, it is really best - and most helpful later on - to deduce a formula for $s(x)$. Say we have the prime factor decomposition $x=p_1^{r_1} \cdot \ldots \cdot p_n^{r_n}$. Then
$$s(x)= \sum_{d|x} d = \prod_{i=1}^{n} \dfrac{N_i}{D_i} = \left( \dfrac{N_1}{D_1} \right)\cdot \ldots \cdot \left( \dfrac{N_n}{D_n} \right)$$
If you find (by counting) those numerators $N_i$ and denominators $D_i$ which only depend on those primes, then the proof that $s(x)$ is multiplicative is really easy. Moreover, you can also tackle problem #3, AND you will have a nice formula for future use!

#### fbs7

You need to quantify your variables: for any / for all or there exist. Otherwise it is not clear what you mean. E.g.

Every divisor $d$ of the form $d=uv$ ... what? If $u,v$ are as given, then $d:=uv$ divides the product? All divisors are of this form? We only know this in case $x$ and $y$ are coprime, but you just said, that you want to consider the general case.
Oh, I thought that was a general result! coming back to that! Thank you!

#### fbs7

Alright, trying my hand at High School 3, and trying to get all my "i" dotted and "t" crossed this time. Doing it the long way - no shortcuts, or I'll end up kicked from the forum and sent back to pre-school! First part, for the sigma of prime decomposition

I - Define $\sigma(x)$ = sum of divisors of $x$, including 1 and $x$, for $x \in N$
--------------------------------------------------------------------------------------------------------
II - Proof that if $p$ is prime, $p > 1$ and $n>=1$, then $\sigma(p^n) = (p^{(n+1)}-1)/(p-1)$

As $p$ is prime, $p>1$ and $n>=1$, the divisors of $p^n$ are {$1, p, p^2, ..., p^n$ }; the sum of these divisors form the sum of a geometric series:

$\sigma(p^n)$ = $1+p+p^2+...+p^n$

this series has $a=1, rate=p, num=n+1$; as $rate>1$, the series has $sum=a*(rate^{num}-1)/(rate-1) = 1*(p^{(n+1)}-1)/(p-1)$, therefore

$\sigma(p^n) = (p^{(n+1)}-1)/(p-1)$
--------------------------------------------------------------------------------------------------------
III - Proof that if $p$ is a prime, $p>1$, $a \in N$, $a>1$, $n>=1$, and $p$ is not a factor of $a$, then $\sigma(p^n*a)=\sigma(p^n)*\sigma(a)$

(a) Define

$z = p^n*a$

(b) Define

P=set of divisors of $p^n=\{1,p,p^2,...,p^n\}$
A=set of divisors of $a = \{a_0,a_1,a_2,...,a_m\}$, with $a_0=1, a_m=a$

(c) For any $u \in N, u>1$, such that $u|z$ and $gcd(u, p^n)=1$; then $u|a$; that's due to Euclid's Lemma; as $u|a$, $u \in A$, therefore $u = a_j$ for some $j$, and $u=p^0*a_j$ for some j

(d) For any $u \in N, u>1$, such that $u|z$ and $gcd(u, p^n)=p^l$ for some $l>0$ ; then necessarily $u=p^l*v$ for an integer $v$, where $gcd(v,p^n)=1$; then for some $k_u \in N, k_u>0$:

$z=k_u*u=p^n*a$
$u=p^l*v$
$z/u=k_u=(p^n*a)/(p^l*v)$
$k_u=(p^n/p^l)*(a/v)$

as $n>l$ and $k_u$ is an integer and $v$ is not a multiple of $p$, then necessarily $a$ is a multiple of $v$, or $v$ is a divisor of $a$:

$v|a$

therefore $v \in A$, what means $v=a_j$ for some $j$, and

$u=p^i*a_j$ for some $i,j$

(e) From (c) and (d), all divisors of z have the form $p^i*a_j$ for some 0<=i<=n, 0<=j<=m.

(f) Therefore the sum of the divisors of z is

$\sigma(z)=\sum_i (\sum_j (p^i*a_j)) = \sum_i (p^i * (\sum_j a_j)) = \sum_i (p^i * \sigma(a)) = \sigma(a)*(\sum_i (p^i)) = \sigma(a)*\sigma(p^n)$
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IV - Proof that $\sigma ( \prod_i {p_i}^{n_i} ) = \prod_i \sigma( {p_i}^{n_i} )$ if each $p_i$ is a different prime $p_i > 1$ and $n_i>=0, 1<=i<=m$

(a) Define

$z = z_1$
$z_1 = \prod_i {p_i}^{n_i}, i>=1,$ that is $z_1 = {p_1}^{n_1} * z_2$
$z_2 = \prod_i {p_i}^{n_i}, i>=2,$ that is $z_2 = {p_2}^{n_2} * z_3$
...
$z_{(m-1)} = \prod_i {p_i}^{n_i}, i>=m-1,$ that is $z_{(m-1)} = p_{(m-1)}^{n_{(m-1)}} * z_m$
$z_m = \prod_i {p_i}^{n_i}, i>=m,$ that is $z_m = {p_m}^{n_m}$

then, as each $p_i$ is a different prime,

$\sigma(z_m) = \sigma({p_m}^{n_m})$
$\sigma(z_{(m-1)}) = \sigma( ({p_{(m-1)}}^{n_{(m-1)}} * z_m ) = \sigma(p_{(m-1)}^{n_{(m-1)}})*\sigma(p_m^{n_m})$
...
$\sigma(z_1) = \sigma(p_1^{n_1} * z_2) = \sigma({p_1}^{n_1})*\sigma(z_2) = \sigma({p_1}^{n_1})*...*\sigma({p_m}^{n_m}) = \prod_i \sigma( {p_i}^{n_i} )$

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V - Proof that $\sigma( \prod_i {p_i}^{n_i} ) = \prod_i (p^{({n_{i}}+1)}-1)/(p-1)$ if each $p_i$ is a different prime $p_i > 1$ and $n_i>=0$, for $1<=i<=m$

(a) Just substitute II in IV

VI - Result: if the prime factorization of $z$ is $z = \prod_i {p_i}^{n_i}$, then $\sigma(z) = \prod_i (p^{({n_{i}}+1)}-1)/(p-1)$

Last edited:

"Math Challenge - March 2019"

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