# Does a solution exist where $p,q$ are odd primes and $p^a - q^b = p^c - q^d$ where $a > c > 1$ and $b > d > 1$

From my thinking so far, there is no solution. Is this an open question or is the answer well known?

If a solution exists, then:

$$p^c(p^{a-c} - 1) = q^d(q^{b-d} - 1)$$

Since $p^c \mid (q^{b-d}-1)$ and $q^d \mid (p^{a-c}-1)$, it follows that:

$$q^{b-d} \equiv 1 \pmod {p^c}$$

and

$$p^{a-c} \equiv 1 \pmod {q^d}$$

Assuming that we want to minimize $b-d$ and $a-c$, using Carmichael's Theorem, it follows that:
$$b-d = \varphi(p^c) = (p-1)p^{c-1}$$ $$a-c = \varphi(q^d) = (q-1)q^{d-1}$$

Now, this looks preposterous to me:

$$p^c - q^d = p^{c+(q-1)q^{d-1}} - q^{d+(p-1)p^{c-1}}$$

where $c,d \ge 2$ and $p,q \ge 3$

If we assume a multiple of $\varphi(p^c)$ and $\varphi(q^d)$, it still looks wrong to me:

$$p^c - q^d = p^{c+u(q-1)q^{d-1}} - q^{d+v(p-1)p^{c-1}}$$

where $c,d \ge 2$ and $p,q \ge 3$ and $u,v \ge 1$

Is there a straight forward way to show this is an impossible equation where $p,q,c,d$ are all positive integers? Is this a tougher problem than it seems?

Thanks very much,

-Larry

Edit: I added additional information to make it clearer why $b-d = \varphi(p^c)$ David is exactly right. If a solution exists, it will be a multiple.

-
I don't follow how you got $b-d=\varphi(p^c)$. I can see that $b-d$ must be a multiple of $\varphi(p^c)$, but why must they be equal? – David Apr 3 '14 at 21:29
if you switch to + signs, 32 + 4 = 27 + 9. with minus signs, 32 - 8 = 27 - 3, i guess that's why you want exponents larger than 1. – Will Jagy Apr 3 '14 at 21:50
I think the way to approach is would be: $p^c(p^{a-c}-1)=q^d(q^{b-d}-1)$ factoring out $p-1$ and $q-1$ gives $$p^c(p-1)\sum_{i=1}^{a-c-1} p^i=q^d(q-1)\sum_{j=1}^{b-d-1} q^j$$ $(p,q)=1$, and WLOG $p>q\Rightarrow (p,q-1)=1\Rightarrow p|\sum_{j=1}^{b-d-1} q^j$. – Stella Biderman Apr 3 '14 at 22:16
In fact, that bottom row can be about $p^c$ not $p$ – Stella Biderman Apr 3 '14 at 22:17
Also, we have that $q|\sum p^i$, as long as $q^d>p-1$ – Stella Biderman Apr 3 '14 at 22:40

As I understand it, the answer may be yes. Bennett shows that for $a^x - b^y = c$ where $a,b,c$ are nonzero integers with $a,b \ge 2$, then the equation has at most $2$ solutions in positive integers $x$ and $y$.