Calculate all $n \in \Bbb N \setminus \{41\}$ such that $\phi(n)=40$?

Question

I'm looking for an $n \in \Bbb N$ for which $\phi(n) = 40$ where $\phi$ is a Euler-Totient Function

I already found one, namely, $n=41$

How the calculate the $n's$?

Answer 1

If $p$ and $q$ and two distinct primes, then $\varphi(pq)=\varphi(p)\varphi(q)=(p-1)(q-1)$.

So if you choose $p=11$ and $q=5$ you're done !

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Now you can search for all solutions. Using essentially the same method as described in the links given above, it is not difficult to prove that the set of solutions for the equation $\varphi(n)=40$ is :

$$S=\{41,55,75,82,88,100,110,132,150\}$$

Answer 2

Note that $$ \phi (55)= \phi(11)\phi(5) = 10\times 4 =40$$ and

$$ \phi (100)= \phi(4)\phi(25) = 2\times 20 =40$$ as well.

That is the function is not one-to-one.

Answer 3

Hints:

Use the multiplicative property of $\phi$: if $n$ and $m$ are relatively prime then $\phi(nm)=\phi(n)\phi(m)$.

$40=2\times2\times2\times5*=2\times2\times10=4\times2\times5*=8\times5*=4\times10=20\times2.$

$*\phi(n)$ cannot be an odd number $> 1.$

If $\phi(n)=2$ then $n=3, 4, $ or $6;$ note that $6$ is not relatively prime to the others.

If $\phi(n)=4$ then $n=5, 8, 10, $ or $12.$

If $\phi(n)=10$ then $n=11$ or $22.$

If $\phi(n)=20$ then $n=25, 33, 44, 50, $ or $66.$

$3\times4\times11; 5\times11, 5\times22, 8\times11, 10\times11, 12\times11; 3\times25, 4\times25, 6\times25, 4\times33; 41, 41\times2$

Answer 4

For $1<n\in \Bbb N,$ if $n$ has exactly $m$ prime divisors then $$n=\prod_{j=1}^m(p_j)^{E(p_j)}$$ where $\{p_1,...,p_m\}$ is the set of all prime divisors of $n$, and each $E(p_j)\in \Bbb N.$ And then $$\phi(n)=\prod_{j=1}^m(p_j-1)(p_j)^{E(p_j)-1}.$$ If $\phi(n)=40$ then

(i). For each odd $p_j$ we have $(p_j-1)|40$ so the only possible values for an odd $p_j$ are $3,5,$ and $11.$

(ii). If $3|n$ then $E(3)=1$ Otherwise $3|3^{E(3)-1}|40, $ implying $3|40.$ Similarly if $11|n$ then $E(11)=1.$

(iii). If $5|n$ then $E(5)\le 2.$ Otherwise $5^2|5^{E(5)-1}|40,$ implying $5^2|40.$

(iv). If $2|n$ then $E(2)\le 4.$ Otherwise $2^4|2^{E(2)-1}|40,$ implying $2^4|40.$

(v). By (i) we have $$n=2^A3^B5^C11^D$$ where $A,B,C,D$ are non-negative integers. By (ii), (iii), and (iv) we have $A\le 4,\, B\le 1,\,C\le 2,\, D\le 1.$

(vi). If $A=4$ then $B=C=D=0.$ Otherwise $2^3|2^{A-1}$ and at least one of the (even) terms $3^{B-1}(3-1),\, 5^{C-1}(5-1),\,11^{D-1}(11-1)\,$ would also occur in the product for $\phi(n),$ implying $2^3\cdot 2=16|40.$

But if $A=4$ and $B=C=D=0$ then $n$ is a power of $2$ so $\phi(n)\ne 40.$

Therefore $A\le 3.$

(vii). If $D=0$ then $5|40=\phi(n)=\phi(2^A3^B5^C)$ which requires $C\ge 2.$ But by (ii) we have $C\le 2,$ so $D=0\implies C=2.$

If $D=1$ then $C\le 1$ otherwise $5^2|5^{C-1}(11-1)|\phi(n)=40,$ implying $5^2|40.$

(viii). This narrows it to $16$ potential cases: $A\in \{0,1,2,3\}$ and $B\in \{0,1\}$ and $(C,D)\in \{(2,0),(1,1)\}.$ Now I will leave it to you to find which of these give $n\ne 41$ and $\phi(n)=40.$