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November 5

Pillai's theorem: $|A{x}^n-B{y}^m|\gg x^{\lambda n}$

For every $x>1,y>1,n>1,m>1$, Pillai's theorem states the following:

• The difference $|A{x}^n-B{y}^m|\gg x^{\lambda n}$ for any λ less than 1, uniformly in m and n.

I wonder if one can prove (e.g. by his theorem) the following claim:

• There exists $x$, such that for every $n>2,m>1$, there exist $a,b$ such that every $A>a,B>b$ satisfy $|A{x}^n-B{y}^m|>(x+2{)}^2$.

If the answer is positive, and $P_x$ is the minimal prime larger than any given $x$, then I also wonder if one can prove (e.g. by his theorem) the following:

• There exists $x$, such that for every $n>2,m>1$, there exist $a,b$ such that every $A>a,B>b$ satisfy $|A{x}^n-B{y}^m|>P_x^2$.

185.24.76.181 (talk) 14:11, 5 November 2021 (UTC)

I'd like to see a precise statement of the theorem, written out with explicit quantifiers. Perhaps I misunderstand the $\gg$ notation, but I doubt that, uniformly,

$|(x^2{)}^n-x^{2n}|\gg \sqrt{x}.$

 --Lambiam 18:31, 5 November 2021 (UTC)

Well, he at least means that for every $x,y$ and every $n>1,m>1$, there exist $a,b$ such that every $A>a,B>b$ satisfy $|A{x}^n-B{y}^m|>x^{\lambda n}$ for any λ less than 1. 185.24.76.176 (talk) 19:23, 6 November 2021 (UTC)

So set $x=y=1,n=m=2,$ and given $a$ and $b$ with the stated property whose existence is promised for these values, set $A=B=\max (a,b)+1.$ Then the lhs of the inequation equals $0,$ while the rhs equals $1.$  --Lambiam 22:57, 6 November 2021 (UTC)

Well, I was wrong with my interpretation. Reading our article about Pillai's theorem, I'm sure he at least meant that for every $x>1,y>1,n>1,m>1$, there exist $a,b$ such that every $A>a,B>b$ satisfy $|A{x}^n-B{y}^m|>x^{\lambda n}$ for any λ less than 1. 185.24.76.176 (talk) 23:26, 6 November 2021 (UTC)

Then set $x=y=2,$ and the rest as before.  --Lambiam 23:55, 6 November 2021 (UTC)

Oh, so weird! Thanks to your comment, now I wonder what our article means - quoting Pillai's theorem. 185.24.76.176 (talk) 10:53, 7 November 2021 (UTC)