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If $d_1 =1, d_2, \cdots, d_k = n$ are the positive divisors of the natural number $n$, show that $(d_ld_2 \cdots d_k)^2 = n^k$.

Seems a confusing question with no place to start with, please help. I see no approach except induction to work. Apart from induction, if some way to bring up a proof, then would be quite fine.

jitender
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2 Answers2

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In this case the induction is just multiplication, so you can "unroll" it as follows:

Let the prime factorization of $n$ be $n=p_1^{a_1}...p_r^{a_r}$. Then the divisors are $D= \{p_1^{b_1}...p_r^{b_r}:0\leq b_i\leq a_i\}$ with $k=|D|=(a_1+1)...(a_r+1)$.

Multiplying them all together with the fact that the "expected" exponent of $p_i$ is $a_i/2$ (*),

$$(\prod_{d\in D} d)^2=\prod_{d\in D} d^2=(p_1^{a_1/2}...p_r^{a_r/2})^{2|D|}=(p_1^{a_1}...p_r^{a_r})^k=n^k$$

(*) You can think of an element in $D$ as a tuple $(b_1,...b_r)$ of exponents, then $D$ is a cartesian product $\{0,1,...a_1\}\times ... \times \{0,1,...a_r\}$. Multiplying elements of D is like adding their exponents, and it's straightforward to show the average tuple is $(a_1/2,...a_r/2)$.

Akababa
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  • Answer to my question is difficult to get from your answer. I can see that you have put powers to the divisors ($b_i$), but if the expression were elaborated then may be better understood from that. I request an example, if possible; else allow me to produce one for vetting. – jitender Feb 03 '18 at 20:48
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    Should I elaborate how I used the exponent laws in the last line? – Akababa Feb 03 '18 at 20:51
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    I explained it a bit more, hopefully that makes it more clear. Try it for some small number like 12. – Akababa Feb 03 '18 at 21:00
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    You forgot 3. The basic idea is that the geometric mean of the divisors is $\sqrt 12$, as you can pair up $d$ and $12/d$. – Akababa Feb 03 '18 at 21:13
  • Thanks for pointing out the error of missing $3$ as a factor. For $12$, the divisors are $1,2,4, 6, 12$, with $d_1=1, d_2=2, d_3=4, d_4=3,d_5=6, d_6=12$. $\prod_{d\in D} (d)^2= (2\cdot 3\cdot 4\cdot6\cdot 12)^2\implies (2\cdot 3\cdot 2^2\cdot 2.3\cdot 2.2.3)^2\implies (2^6\cdot 3^3)^2 = 12^6 \implies (2\cdot \sqrt{3})^2 = 12.$ I have arrived at logic, by example;but the understanding is nil, i.e., why it occurs. Your answer also is based on an observed fact, but the logic is not clear. – jitender Feb 03 '18 at 21:24
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    If you sort the divisors $1,2,3,4,6,12$, you can pair the first and last, second and second last, etc. $112,26,3*4$ and they all equal 12. You get $k/2$ such pairs, so multiplying them together gets $12^{k/2}$ – Akababa Feb 03 '18 at 21:29
  • So, it is a property of divisors. Say for $12$, there are two trivial divisors ($1$ and itself), and (I am hypothesizing) rest an even number of, i.e. $2+2^n$, here $n=2$. Is there any pattern to the number of non-trivial divisors (just like $2^2$ for $12$). Do the simple divisibility rule applies, i.e. for a number $n$, if $d$ is a divisor, then $\frac{n}{d}$ is also a divisor. But, the proof is needed. I mean for odd numbers, will get odd divisor pair, likewise for even. That would explain for even no. of divisors, but not about how to find an expression for number of non-trivial divisors. – jitender Feb 03 '18 at 21:40
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Let n =1.n =a.b =c.d =e.f =....... = x.y be all possible ways of writing n where 1,n,a,b,c,d, .....,q,r,.......,x,y represent k factors (even number) in all. (this is true even if n may be a perfect square where say n=q.r and q=r)

The k factors are organized into k/2 pairs of divisors such that the product of every pair is equal to n.

so the product of all divisors will be equal to (n)^ (k/2)

so the square of the product of all divisors will be equal to (n)^k

prashanth rao
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  • Please elaborate how the $k$ factors may be organized into $\frac{k}{2}$ pairs of divisors, s.t. product of every pair of primes is equal to $n$. – jitender Feb 03 '18 at 20:46