0
$\begingroup$

How do you bound the exponent of $x^2+1=y^p$ for $p$ a prime exponent using linear forms in logs?

So far I have $(x-i)(x+i)=y^p$ which are coprime and hence $x+i=(a+ib)^p$.

Now how do I get a linear form in logs so that I can find an upper bound on $p$?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Catalan%27s_conjecture $\endgroup$
    – Will Jagy
    Commented Jun 15, 2012 at 19:43
  • 1
    $\begingroup$ not what link is supposed to mean, I'm not asking for a full proof of cataln's conjecture but consider this particular e.g. $\endgroup$
    – Kale
    Commented Jun 15, 2012 at 20:34
  • $\begingroup$ perhaps you can give some references that use the techniques you want to apply. $\endgroup$
    – Will Jagy
    Commented Jun 15, 2012 at 23:53
  • $\begingroup$ You can look up the work of Tijdeman for this kind of argument. But why would you not want to use a known theorem? $\endgroup$
    – Felipe Voloch
    Commented Jun 16, 2012 at 0:05

1 Answer 1

Reset to default
1
$\begingroup$

If $x+i=(a+ib)^p$, then $x-i=(a-ib)^p$, $2i=(a+ib)^p-(a-ib)^p=2ib$(smth), $b=\pm 1$, $1=\pm 1\mp \binom{p}2a^2\pm \binom{p}4 a^4+\dots$. Considering this modulo $a^2$ we see that signs start from plus, that is, $\binom{p}2a^2\pm \binom{p}4 a^4+\dots=0$. Next, $\binom{p}{2k}=\binom{p}2\binom{p-2}{2k-2}/\binom{2k}2$ and dividing by $\binom{p}2a^2$ we get $$ 1=\sum_{k\geqslant 2} \pm \binom{p-2}{2k-2}\frac{a^{2k-2}}{k(2k-1)}. $$ If $a$ has some prime divisor $q$, then RHS has positive $q$-adic valuation, since $k(2k-1)$ can not be divisible by $q^{2k-2}$. If $a=\pm 1$, then we know what is $(a+i)^p$ and it is not $x+i$.

This is possibly not what you were asking for, but at least it solves this case of Catalan conjecture very quickly.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .