Solving $\phi(n)=84$

Question

Ok, I really need some help understanding this because either my brain isn't working at the moment or I'm breaking math and I have a striking suspicion that one of those is more likely.

Anyways, here's my process. WolframAlpha tells me there are 12 integer solutions, they are all $^+_-n$ so let's say there are 6 of them. They are as follows: $129, 147, 172, 196, 258, 294$

Ok, so here's how I attempted to solve it. $n$ looks like this ${\prod_{i=1}^kp_i^{\alpha_i}}$. And $\phi(n)=n\prod_{p\vert n}(1-\frac{1}{p})$. So I plug $n$ in and get an equation like this:

$\prod_{i=1}^k p_i^{\alpha_i-1}(p_i-1)=84$

Now, solving for $n$ SHOULD be equivalent to solving for $p_i, \alpha_i$, right?

So, I started going through possible values of $k$ and I'm interested in the case of $k=2$.

This means that $n$ has two prime factors, so it looks like $p_1^{\alpha_1}p_2^{\alpha_2}$, and the equation is:

$p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)=84$

Ok. Here's my issue. Factoring $84$, we get $2\times2\times3\times7$

So I try to see a pattern. The first solution I came up with is $p_1=7, \alpha_1=2, p_2=2, \alpha_2=2$. Then I followed the same logic that led me to that solution but didn't find any more so I concluded that the other solutions must be as higher values of $k$. But then I factored $172$ and saw that it also had only two primes. It also fits the pattern. So I realized that I could get ones in that product as a result of $p^0$ and following that pattern, filling in the remaining factors, I found the $129$ solution.

No more, the next one must be at $k=3$. NOPE. $147$ also factors into only 2 primes. So now I'm really at a miss. Is there a more effective way to iterate over all of these combinations?

Answer 1

Look at 7. Two possibilities:

• $7|(p-1)$ with $p|n$: hence $p = 14k+1 \in \{ 29, 43\}$ (the others are too big).
• If $p=29$, $\phi(n) = 28\times 3 = (29-1)(3)$. 3 has to be some $(p-1)p^{a-1}$ but $p=4$ is not prime.
• If $p=43$, you have the solution $\phi(n) = 42\times 2\implies n = 43\times 3$ or $n = 43\times 2^2$.
• Otherwise: $7|n$. $\phi(n) = 6\times 7\times 2\implies n= 7^2\times 3$ or $n= 7^2\times 2^2$.

I have used that $\phi(n) = 2\implies n\in\{3,4\}$.

Answer 2

Here is a technique you can use to solve these kinds of problems that doesn't need much intuition. Firstly, you need a lemma.

Lemma: If $n$ is odd, then $\phi(n) =\phi(2n)$.

If the prime factorisation of $n$ is $n = \prod_{i=1}^m p_i^{\alpha_i}$, then $\phi(n) = \prod_{i=1}^m p_i^{\alpha_i-1}(p_i-1)$. So $p_i - 1 \mid 84$. Listing out factors of 84, the possible values $p_i-1$ can take are:

$$p_i-1 = 1,2,4,6,7,12,14,21,42,84.$$

Then, $$p_i = 2,3,5,7,8,13,15,22,43,85,$$ of which $8,15,22,85$ are not prime. So possible values of $p_i$ can take are: $$p_i = 2,3,5,7,13,43.$$

Now condition on the highest prime occurring in the prime factorisation of $n$.

$43$ is the highest prime dividing $n$: Then $\phi(n) = 42*a, a=2$. Then we only need to hunt for something to contribute a factor of $2$ for $\phi(n)$. This can come from $2^2$ or $3$. This gives $n = 43*3$ or $n=43*2^2$ respectively. With the lemma, $\phi(2*43*3)=\phi(3*43)$. This gives $3$ solutions, $n= 129,258,172$.

$13$ is the highest prime dividing $n$. $\phi(n) = 12 *a , a=7$. We need to hunt for a term contributing $7$. $7$ can either come from one of the $p_i-1$'s or $p_i$'s. It is the latter because $p_i-1 = 7 \implies p_i = 8$, which is not prime. Yet, having $7 \mid n$ would imply $6 *12 \mid \phi(n)$, contradiction. So $13$ cannot be the highest prime dividing $n$.

$7$ is the highest prime dividing $n$. $\phi(n) = 6*a, a=14$. We need to hunt for a term contributing $14$. Yet all primes dividing $n$ are going to be smaller than $7$ already, so the only option for $\phi(n)$ to incur a factor of $7$ would be for the power of $7$ in the prime factorisation of $n$ to be $\geq 2$. We can further see that it has to be exactly $2$. This means that $7^2 \mid n$ and $\phi(n) = 7*6*a, a=2$. We still need a factor of $2$, which would come from $3$ or $2^2$. So $n = 7^2 * 3$ or $n = 7^2 * 2^2$. From the lemma, $n = 2*7^2 *3$ is a viable option as well. So $n = 147,294, 196$.

$5$ is the highest prime dividing $n$. Then we need to hunt for a term contributing $21$. Yet this implies $7 \mid n$, but now the primes are too `small' to contribute a $7$, contradiction.

Similarly, for $3$ and $2$ being the highest prime dividing $n$, it will always imply that $7 \mid n$. So there are no more answers.

Answer 3

  84         129 = 3 * 43
  84         147 = 3 * 7^2
  84         172 = 2^2 * 43
  84         196 = 2^2 * 7^2
  84         258 = 2 * 3 * 43
  84         294 = 2 * 3 * 7^2

I see, you had those.

I suppose what I would emphasize is this: $\phi$ is multiplicative. So it is determined by its behavior on prime powers. Next, $\phi (p^k)$ is always divisible by $p-1.$ So, we are required to have $(p-1) | 84,$ so out of that list of numbers with $\phi(n) | 84,$

   1           2 = 2
   2           3 = 3
   2           4 = 2^2
   4           5 = 5
   2           6 = 2 * 3
   6           7 = 7
   4           8 = 2^3
   6           9 = 3^2
   4          10 = 2 * 5
   4          12 = 2^2 * 3
  12          13 = 13
   6          14 = 2 * 7
   6          18 = 2 * 3^2
  12          21 = 3 * 7
  12          26 = 2 * 13
  12          28 = 2^2 * 7
  28          29 = 29
  12          36 = 2^2 * 3^2
  12          42 = 2 * 3 * 7
  42          43 = 43
  42          49 = 7^2
  28          58 = 2 * 29
  42          86 = 2 * 43
  42          98 = 2 * 7^2
  84         129 = 3 * 43
  84         147 = 3 * 7^2
  84         172 = 2^2 * 43
  84         196 = 2^2 * 7^2
  84         258 = 2 * 3 * 43
  84         294 = 2 * 3 * 7^2

the primes allowed are 2,3,5,7,13,29,43, I guess that's it. Some evident prime powers that need to be considered as alternatives to just primes. Oh, 13 does not get used, because then you need $\phi(something) = 21,$ and $\phi$ is never odd unless it is exactly $1.$ Again, 9 does not get used, i don't think there is any number with $\phi = 14,$ don't remember why, but people ask that here fairly often so you can find an answer on MSE.. The bit about odd is fairly easy, if there is any odd prime factor $p,$ then $(p-1)$ is even. So all that is left is powers of $2.$