Proof that $\phi (n) \neq 50$? Like the question states, I'm curious how to go about doing this.
I found so far that if such an n does exist, then $n \in \Big[51,219 \Big]$, but I'm not sure if I took the right route with this one.
Thanks for the tips.
 A: $\phi$ is multiplicative with $\phi(p^d) = (p-1) p^{d-1}$ for primes $p$.
Each odd prime power, and each power of $2$ greater than $2$ itself gets at least one factor of $2$ from this, and $50$ is
divisible by $2$ but not $4$, so there can be only one of those, i.e. 
$n$ is either $2$ (which is easy to rule out) or a prime power.  The only other prime divisor of $50$ is $5$.  But $n$ can't be a power of $5$, since again that would contribute a factor of $4$.  Thus we must have $50 = p-1$ with $d=1$.
But $51$ is not prime...  
A: If $\phi(n) = 50$, then $50$ is the product of numbers of the form $(p - 1)p^\alpha$, where $p$ is prime and $\alpha \geq 0$. But $50 = 2 \times 5 \times 5$: although $2$ is $1$ less than a prime ($3$), neither $5$ nor $25$ is. We could try $10$, which is $1$ less than $11$, but that leaves us with a "loose" $5$.
To be better convinced, try solving $\phi(n) = 60$. One answer should immediately suggest itself: $n = 61$. Another answer is $n = 93$, this comes from the observation that $60 = 2 \times 30$. There may be more, you get the idea.
A: Prove the following easy fact: if $n$ has $k$ different prime factors, then $2^k$ divides $\phi(n)$ (hint: what's the formula for $\phi(n)$ in terms of prime factors of $n$?). It follows that $n$ can only have 1 distinct prime factor. Which means that $n=p^k$ for a prime $p$. But $\phi(p^k)=p^k(1-1/p)$. 
So 
$$p^{k-1}(p-1)=50,$$
and this has no solution for any $p,k$ since $50=2\cdot 5^2$.
