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 $1 \leq j,k \leq t$

Solution 1:

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 $\phi(x) = m$.