If $p$ is a prime and $a,b,c \in \mathbb{N}$, then does $p^a-p^b =p^c$? Let $p$ be a prime, then for $a,b,c \in \mathbb{N}$, can we definitively assert that there exists no $c \in \mathbb{N}$ so that
$$p^a-p^b =p^c$$
In other words, can we assert that the difference of powers of $p$ will never be a power of $p$?
 A: This is easy to think about in base $p$ instead of base ten. Suppose $p^a=p^b+p^c$. Then in base $p$:
$$1\overbrace{0\cdots0}^a=1\overbrace{0\cdots0}^b+1\overbrace{0\cdots0}^c$$
Suppose that $b\neq c$. Then the right side is $1\overbrace{0\cdots0}^{b-c-1}1\overbrace{0\cdots0}^c$, clearly not equal to $1\overbrace{0\cdots0}^a$.
If $b=c$ and $p\neq2$, the right side is $2\overbrace{0\cdots0}^b$, clearly not equal to $1\overbrace{0\cdots0}^a$.
So this is only possible when $b=c$ and $p=2$. In which case $a$ must be $b+1$. Such a situation is noted in the comments.
A: Clearly $p^a-p^b$ can only be a positive integer if $b<a$; thus observe that $p^a-p^b=p^b(p^{a-b}-1)$.
This can only be some power of $p$ ($p^c$) if $p^{a-b}-1$ is a power of $p$. That is, two powers of $p$ ($p^c,p^{a-b}$) are consecutive integers.
The only such case is $1=2^0,2=2^1$, so that $p=2$ (and $c=0,a-b=1$). Any $a,b$ such that $a-b=1$ will work, as $2^a-2^{a-1}=2^{a-1}$.
A: Here's a solution using properties of the p-adic absolute value, just to give a bit of a simple demonstration for interest and fun.
Without loss of generality we can assume $a>b$. We can then take the p-adic absolute value of both sides of the equation,
$$p^a-p^b=p^c$$
$$|p^a-p^b|_p=p^{-c}$$
By the ultrametric inequality's strong property, which forces an equality:
$$|p^a-p^b|_p = \max(|p^a|_p, |p^b|_p) = p^{-b}$$
Comparing with the preceding equation, this makes $b=c$. So going back to our original equation it becomes:
$$p^a=2p^b$$
Obviously the only way we can satisfy this equation is if $p=2$, and so we get, $a=b+1$. Therefore all solutions are of the form:
$$2^{t+1}-2^t=2^t$$
