# If $n_j = p_1\cdot \ldots \cdot p_t - \frac{p_1\cdot \ldots \cdot p_t}{p_j}$, then $\phi(n_j)=\phi(n_k)$ for $1 \leq j,k \leq t$

Show that if $p_1,\ldots p_t$ are the first $t$ prime numbers, and $n_j = p_1\cdot \ldots \cdot p_t - \frac{p_1\cdot \ldots \cdot p_t}{p_j}$, then $\phi(n_j)=\phi(n_k)$ for $1 \leq j,k \leq t$ and conclude that the equation $\phi(x)=m$ has infinitely many solutions. Here $\phi(\cdot)$ is the Euler Totient function.

I am really stuck on this one. First of all $p_j \nmid n_j$ because even though $p_j$ divides the first term in $n_j$ it does not divide the second. Therefore $gcd(n_j,p_j)=1$. However, can we use this to prove the theorem? Thanks for any help!

EDIT At first the whole expression for $n_j$ seemed really confusing to me. I think it can be rewritten as: $$n_j = \left(1-\frac{1}{p_j}\right) \prod\limits_{i=1}^{t} p_i = (p_j-1)\left(\prod\limits_{i=1}^{j-1}p_i \right)\cdot \left(\prod\limits_{i=j+1}^{t}p_i \right)$$ Maybe this helps!

Let $\phi$ denote the Euler totient function. Let $d$, $n$ be positive integers with $d$ dividing some power of $n$, that is, every prime divisor of $d$ also divides $n$. Then $\phi(dn) = d\,\phi(n)$. As $N = p_1\cdots p_t$ is the product of the first $t$ prime numbers, every prime factor of $p_j-1$ still divides $N/p_j$. Therefore, $$\phi(n_j) = \phi\left((p_j - 1)\,\frac{N}{p_j}\right) = (p_j - 1)\,\phi\left(\frac{N}{p_j}\right) = \phi(p_j)\,\phi\left(\frac{N}{p_j}\right)$$ As $\phi$ is multiplicative and $N/p_j$ is coprime to $p_j$, we get $$\phi(n_j) = \phi(N)$$ which implies the statement $\phi(n_j) = \phi(n_k)$.
Nevertheless, $\phi(x) = m$ with given $m$ has only finitely many solutions for $x$. Too see this, consider the number of prime factors of $x$, counted with multiplicity. Let us denote this $\Omega(x)$. You can deduce from the product formula for $\phi(x)$ that $\Omega(\phi(x)) \geq \Omega(x) - 1$ (the $-1$ is for the possible prime divisor 2). Therefore $$\Omega(x) \leq \Omega(m) + 1$$ and since the greatest prime divisor of $x$ cannot exceed $m+1$, we can easily estimate an upper bound for $x$ as $$x \leq (m+1)^{\Omega(m)+1}$$ We conclude that there can only be finitely many $x$ with $\phi(x) = m$.
Call $N = p_1 p_2 \ldots p_t$ for compactness. Then: $$n_j = N - N / p_j = (p_j - 1) N / p_j$$
As the totient is multiplicative and never zero: $$\phi(n_j) = \phi(p_j - 1) \phi(N) / \phi(p_j) = \phi(p_j - 1) \phi(N) / (p_j - 1)$$ If it was $\phi(n_j) = \phi(n_k)$, then it would be: $$\phi(p_j - 1) / (p_j - 1) = \phi(p_k - 1) / (p_k - 1)$$ But: \begin{align*} \phi(5 - 1) / (5 - 1) &= 2 / 4 = 1/2 \\ \phi(7 - 1) / (7 - 1) &= 2 / 6 = 1/3 \end{align*} Too nice to be true :-(
• Hey thanks for your answer. I think the question is meant to mean that there exists $j$ and $k$ smaller than $t$ such that $\phi(n_j)=\phi(n_k)$. For instance in the example you gave $\phi(3)/(3-1) = 1/2$. So it does not need to hold for all $j$ and $k$ smaller than $t$. – Slugger May 3 '13 at 23:18
• @TeunVerstraaten OK, then $p_j = 3$ and $p_k = 5$ will work for OP's needs, for any $t$ – vonbrand May 3 '13 at 23:55
• Yeah but $j$ needs to specified beforehand. Thats why in the question I used $n_j$. And then you show that some $k$ exists to satisfy the condition. – Slugger May 4 '13 at 0:04
• @vonbrand: $\phi$ is multiplicative, but not completely multiplicative. – ccorn May 4 '13 at 1:38