I have a question about very early argument in the proof of Thereom 1.1.

Theorem 1.1 of On Some Exponential Equations of S.S. Pillai states that if $a,b,c$ are nonzero integers with $a,b \ge 2$, then the equation:

$$a^x - b^y = c$$

has at most two solutions in positive integers $x$ and $y$.

The first step in the proof is to assume that three solutions ($x_i,y_i$) exist in positive integers where:

$$x_1 < x_2 < x_3$$ and $$y_1 < y_2 < y_3$$

The part that I am having trouble understanding relates to what should be a simple argument by contradiction:

  • Assume that gcd($a,b$) $> 1$
  • Then, there exists a prime $p$ such that $p \mid a$ and $p \mid b$

Now here's is the step that I am misunderstanding:

  • The claim is made that $p$ has ord$_p{a} = \alpha \ge 1$ and $p$ has ord$_p{b} = \beta \ge 1$ since:

$$a^{x_i}(a^{x_{i+1}-x_i} - 1) = b^{y_i}(b^{y_{i+1}-y_i}-1)$$

But I thought that ord only applied when gcd($p,a$)$=1$ and gcd($p,b$)$=1$. So, I am confused by the statement.

Then, based on the above, it is concluded for $i=1,2$ that:

$$\alpha{x_i} = \beta{y_i}$$

If someone could explain what ord means in this circumstance and how it leads to the conlusion of $\alpha{x_i} = \beta{y_i}$, I would greatly appreciate it.



  • $\begingroup$ Larry, could you please email me something, the gmail address is most convenient, see ams.org/cml if you don't have it. I cannot seem to find an address for you. The short version is that Gerhard would like to get in touch with you. $\endgroup$ – Will Jagy Jul 9 '14 at 21:18

This is the other most common meaning, the $p$-adic valuation. It, $\operatorname{ord}_p a,$ is just the highest exponent, $e,$ of $p$ such that $$ p^e | a.$$ In Gouvea's book this is written $\nu_p a,$ in other books $v_p a,$ the letter $v$ being convenient for the English word valuation. Alright, holding Gouvea's book, page 25, he also uses letter $v.$

  • $\begingroup$ Thanks very much! I thought that there had to be another meaning of the term. $\endgroup$ – Larry Freeman Apr 7 '14 at 3:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.