Question

Show that if n is a positive integer then
Question below:

Answer 1

\[(\sum_{d|n}^{} \tau(d) )^2 = \sum_{d|n}^{} \tau(d)^3 \]

Answer 2

i'm unfamiliar with the notation: d|n care to elaborate?

Answer 3

divides?

Answer 4

yea its the divisor

Answer 5

so taking the summatory function of tau then sqauring it is equal to the summatory of tau cubed

Answer 6

im not sure i understand how the whole expression on the left can be squared. did you just mean to square tau(d)^2 ?

Answer 7

nope its written correctly the lhs is completely squared

Answer 8

interesting..

Answer 9

this is supposed to be a hard problem so its gonna take a while to figure out lol

Answer 10

if it wasn't so late, i'd have more enthusiasm ;) im a little fuzzy right now. ill come back to this tomorrow if you dont already have an answer by then

Answer 11

oh thats alright thanks for looking/trying at the problem ^_^

Answer 12

what is the tau function? :O

Answer 13

the tau function is number of divisors

Answer 14

\[\tau(n) = \sum_{d|n}^{} 1\]

Answer 15

ic

Answer 16

so the question becomes \(\Large\left(\sum\limits_{d|n}\sum\limits_{k|d}1\right)^2=\sum\limits_{d|n}\left(\sum\limits_{k|d}1\right)^3\)?

Answer 17

i think it should be like this:

\[(\sum_{d1|n}^{} \sum_{d2|n}^{} 1 )^2 = \sum_{d1|n}^{}(\sum_{d2|n}^{}1)^3\]

Answer 18

but d1 is the subject of the second Sigma...

Answer 19

d1 is the divisors of the outer sum, d2 is the divisors for the sum of 1
so you sum all d2 first then you pick out d1 of d2

Answer 20

yes, d2 < d1 i think

another approach is to mess with \(\tau\)and \(\sigma\) directly :-

say, \(n = p_1^{k_1}p_2^{k_2}... p_r^{k_r}\) is the prime factorization of \(n\)

Answer 21

since, \(\tau\) is a multiplicative funciton, SUM of of \(\tau\) 's is also a multiplicative function.

so it is sufficient to prove when \(n = p^k\)

Answer 22

but the second Sigma is not referring to the number of divisors of \(n\), but the number of divisors of \(d\)

Answer 23

discovery:
\(\Large n=10\rightarrow(1+2+2+4)^2=1^3+2^3+2^3+4^3\)
\(\Large n=15\rightarrow(1+2+2+4)^2=1^3+2^3+2^3+4^3\)
Both has the same numbers... (this may seem dumb?)

Answer 24

interesting, but 4 is not a divisor of 10 rihgt ?


the proof shrinks to proving for a single prime power \(n = p^k\)
then, +ve divisors of \(n\) : \(d = {p^0, p^1, p^2, .... p^k}\)

should be easy to prove

Answer 25

4 is not a divisor of 10, but 4 is the number of divisors of 10

Answer 26

gotcha :)

Answer 27

\(d ~~~~= {p^0, p^1, p^2, .... p^k} \)
\(\tau(d) = 1, 2, 3, 4, .... k+1\)

Answer 28

p is prime?

Answer 29

\(\sum \limits_{d|n}^{} \tau(d) = 1+2+3+....+k+1 \)

Answer 30

just wonder why is it sufficient to say its sufficient to prove when n = p^k
yes its multiplicative why does that imply that you can prove it using prime factorizations

Answer 31

yup p powers are the prime factorization

Answer 32

but it's not prime factorization, it's factorization

Answer 33

cuz, \(\tau (p^kp^lp^m) =\tau (p^k) \tau (p^l)\tau (p^m)\)

if u prove for one prime power, other combinations will follow.

Answer 34

oh alright i get it i see what your doing now

Answer 35

also, \(\tau\) is multiplication implies that any sum of \(\tau\) is also multiplicative
so it all reduces to proving single prime power factorization

Answer 36

*multiplicative

Answer 37

Factors of \(p^k\): 1, \(p^1\),\(p^2\), ..., \(p^k\)
Taus of factors of \(p^k\): 1, 2, 3, ..., k+1

Answer 38

lets see so if i prove over prime powers then
\[(\sum_{d|n}^{}\tau(d))^2 = (\sum_{d|p^k}^{}\tau(p^k))^2 = \sum_{i =1}^{k} (i+1) = \sum_{i =1}^{k+1} (i) = (\frac{ (k+1)(k+2) }{ 2 })^2\]

Answer 39

So that's the same as proving \(\Large\left(\sum\limits_{n=1}^{k+1}n\right)^2=\sum\limits_{n=1}^{k+1}n^3\)

Answer 40

which is common sense lol.

Answer 41

looks good to me

Answer 42

alright thanks for your help everyone much appreciated :D

Answer 43

i will obtain not any credit :)
May all credit go to @ganeshie8

Answer 44

very good problem ! thanks for sharing wid us :)

Answer 45

haha your welcome xD