4

Consider $a$ and $b$, both positive integers. Is it true that $a^a$ divides $b^b$ implies $a$ divides $b$?

I can't seem to figure out this proof. My intuition is to use the fundamental theorem of arithmetic to break each number into its prime components, however I haven't been able to come up with a solution. Any help would be appreciated.

InsideOut
  • 6,883
njszym
  • 75
  • 9
  • 2
    Here is an exact duplicate. Together with plenty of answers containing just examples and/or elaborations on the rather obvious fact that for a counterexample it is necessary that $a$ is divisible by a square of a prime as well as explanations that when $a$ is not square-free you can cook up a suitable $b$ and get a counterexample. – Jyrki Lahtonen Jan 15 '17 at 07:45

3 Answers3

11

This is not true in general. The fact that $4^4$ divides $10^{10}$ would be the smallest counterexample, I think.

So why does the proposition fail to hold? For any integer $n$, $\operatorname{rad}(n)$ (for "radical") is the number with the same primes in its prime factorisation as $n$, but all powers are $1$. For instance, $\operatorname{rad}(12) = 6$ and $\operatorname{rad}(98) = 14$. Then we do have "$a^a$ divides $b^b$ implies that $\operatorname{rad}(a)$ divides $\operatorname{rad}(b)$", i.e. every prime that appears in the prime factorisation of $a$ appears in the prime factorisation of $b$. However, we have no way of controlling, for any given of those primes, that the power of that prime is smaller in $a$ than in $b$. We just know that for any given of those primes, $a$ times its exponent in $a$ is less than $b$ times its exponent in $b$. Thus, if $b$ is large enough, and just contains the right primes, we get $a^a$ divides $b^b$, no matter how many times those primes divide $b$.

Arthur
  • 199,419
  • 1
    is there a disproof without a counter example? – SAJW Jan 15 '17 at 01:46
  • 4
    @Socrates Why on earth would you want a disproof without a counter example? – Erick Wong Jan 15 '17 at 02:14
  • @ErickWong why not? If it is possible it would be nice. If not, so be it. – SAJW Jan 15 '17 at 02:26
  • 2
    @Socrates I'm not convinced that a disproof without a counterexample would be nice at all. Now, a procedure for generating counterexamples, like Henry's, is interesting, but that's more counterexamples, not a way to avoid them entirely. – David K Jan 15 '17 at 02:38
  • 2
    @Socrates This objection is not that uncommon. I've heard it many times and I have the impression that some students tend to think that it looks "less mathematical" to do it this way, but that is not the case. Providing a counter-example is the main way of disproving a given statement. However other approaches can be very useful in that they might illuminate better why it fails to be true (for example it might be that there is only one single counter-example), but at the end of the day the counter-example as a disproof is the most elegant and simple way of doing it. – Winther Jan 15 '17 at 02:49
  • @Arthur Great! Thank you for the answer. Seems I should have spent more time checking numbers for a counter-example – njszym Jan 15 '17 at 03:03
  • @Socrates I added a little extra, if that helps. – Arthur Jan 15 '17 at 07:02
5

Exceptions will be of the following form:

Suppose the prime factorisation of $a$ is $a=p_1^{e_{a_1}}p_2^{e_{a_2}}\cdots p_n^{e_{a_n}}$ with all the $p_i$ distinct primes and with all the $e_{a_i}\ge 1$ and at least one $e_{a_i}\ge 2$. Then

  • $b$ is divisible by $p_1p_2\cdots p_n$ and can be written as $b=c p_1^{e_{b_1}}p_2^{e_{b_2}}\cdots p_n^{e_{b_n}}$ with $c$ coprime to $a$ and all the $e_{b_i}\ge 1$
  • for at least one $i$ there is $e_{b_i}\lt e_{a_i}$ so $a$ does not divide $b$
  • for all $i$ there is $ae_{a_i}\le be_{b_i}$ so $a^a$ divides $b^b$

So reversing this to generate all possible exceptions

  • choose a positive number $n$ of distinct primes $p_1,p_2,\ldots,p_n$
  • choose the same number $n$ of positive integers $e_{a_1},e_{a_2},\ldots,e_{a_n}$ not all $1$
  • choose the same number $n$ of positive integers $e_{b_1},e_{b_2},\ldots,e_{b_n}$ where for at least one $i$ you have $1 \le e_{b_i}\lt e_{a_i}$
  • find a number $c$ coprime to $p_1p_2\cdots p_n$ and at least $p_1^{e_{a_1}-e_{b_1}}p_2^{e_{a_2}-e_{b_2}}\cdots p_n^{e_{a_n}-e_{b_n}} \displaystyle \max_i\left(\frac{e_{a_i}}{e_{b_i}}\right)$
  • take $a=p_1^{e_{a_1}}p_2^{e_{a_2}}\cdots p_n^{e_{a_n}}$ and $b=c p_1^{e_{b_1}}p_2^{e_{b_2}}\cdots p_n^{e_{b_n}}$
Henry
  • 157,058
  • 2
    Arthur's solution takes $n=1$ (the smallest possible positive integer), $p_1=2$ (the smallest possible prime), $e_{a_1}=2$ (the smallest possible positive integer given $n=1$ and at least one $e_{a_i}\ge 2$ ), $e_{b_1}=1$ (the only possibility given $n=1$ and $e_{a_1}=2$ and at least one $e_{b_i}\lt e_{a_i}$), and $c=5$ (the smallest number coprime to $2$ and at least $2 \times 2$), making $a=2^2=4$ and $b=5\times 2^1=10$ – Henry Jan 15 '17 at 02:35
1

If $ a^a\mid b^b $, then for every prime $ p$ such that $ p\mid a$ we must have $ p\mid b $ because $ p\mid b^b $ implies $ p\mid b $. Thus we can write the prime factorizations of $ a$ and $ b$ as $a=p_1^{\alpha_1}\cdots p_r^{\alpha_r} $ and $ b=p_1^{\beta_1}\cdots p_r^{\beta_r}\cdots $. Therefore $$ a^a=p_1^{a\alpha_1}\cdots p_r^{a\alpha_r} ,$$ $$ b^b=p_1^{b\beta_1}\cdots p_r ^{b\beta_r}\cdots $$

So if $ a^a\mid b^b $ we must have $ a\alpha_i\le b\beta_i $, for every $1\le i\le r$, but this doesn't mean that $\alpha_i\le \beta_i $ for every $ i $. It could happen that for some $ i$, $ \alpha_i> \beta_i $ , which would imply that $ a \not\mid b $.

As a example we have the counterexample given in the first answer: $4^4\mid 10^{10} $, but $4\not\mid 10$. In this case $ r=1, p_1=2, \alpha_1=2$ and $\beta_1=1$. So we have $4\cdot 2\le 10\cdot 1$, but $2> 1$.

Xam
  • 6,119