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The theorem is Theorem 1.1 from Michael A. Bennett in his "On Some Exponential Equations of S.S. Pillai".

Here is the statement of the theorem:

Theorem 1.1. If $a,b,c$ are nonzero integers with $a,b \ge 2$, then the following equation has at most two solutions in positive integers $x$ and $y$.

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

The proof begins by assuming that three such solutions ($x_i,y_i$) exist in positive integers where:

$$x_1 < x_2 < x_3 \,\,\text{and}\,\,y_1 < y_2 <y_3$$

I have a problem understanding one of the intermediate steps when Bennett is trying to show that we can assume that gcd($a,b$)$=1$. Here's the argument:

  • Assume that gcd($a,b$)$>1$
  • There exists a prime $p$ dividing $a$ and $b$ such that ord$_p{a} = \alpha \ge 1$ and ord$_p{b} = \beta \ge 1$

Note: Thanks to Will Jagy's answer to a previous question, I understand that ord$_p{a}=\alpha$ means that $p^{\alpha} \mid a$ but $p^{\alpha+1} \nmid a$.

Now, it's the point after here which I am not understanding. Here's the claim:

  • Since $a^{x_i}(a^{x_{i+1}-x_i}-1)=b^{y_i}(b^{y_{i+1}-y_i} - 1)$, it follows that $\alpha{x_i}=\beta{y_i}$, for $i=1,2$

But this does not seem correct to me. For example, in the same paper, Bennett notes that:

$$6-2 = 6^2 - 2^5$$

In this case, as I understand it, $p=2$, $\alpha=2$, $\beta=1$, $x_1=1$, $y_1=1$, but clearly:

$$2\cdot{1} \ne 1\cdot{1}$$

What am I misunderstanding? Any help would be greatly appreciated.

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  • $\begingroup$ But if $p$ divides $a$ then it doesn't divide $a^{x_{i+1}-x_i}-1$. That is all is being used. $\text{ord}_p(a^{x_i}(a^{x_{i+1}-x_i}-1))=\text{ord}_p(a^{x_i})+\text{ord}_p(a^{x_{i+1}-x_i}-1)=\text{ord}_p(a)x_i+0$. Likewise for $b^{y_i}(b^{y_{i+1}-y_i}-1)$. $\endgroup$
    – user141267
    Apr 8, 2014 at 19:33

1 Answer 1

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Actually, it's just that $$\alpha = 1$$ So your counter-example does not work.

Now, the proof of what seemed unclear to you:

Since $$ p | a^{x_{i+1} - x_{i}} $$ It follows that $$ p \nmid (a^{x_{i+1} - x_{i}} - 1)$$ Then $$ \operatorname{ord}_p a^{x_i}(a^{x_{i+1} - x_{i}}) = \operatorname{ord}_p a^{x_i} = \alpha x_i$$ Similarly $$ \operatorname{ord}_p b^{y_i}(b^{y_{i+1} - y_{i}}) = \operatorname{ord}_p b^{y_i} = \beta y_i$$ And since these first numbers are equal $$\alpha x_i = \beta y_i.$$

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  • $\begingroup$ Thanks. I appreciate the explanation. Now, super clear. $\endgroup$ Apr 8, 2014 at 20:09

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