In 1976 Tijdeman proved that the Catalan equation $$ x^{p}-y^{q}=1 $$ has finitely many solutions in integers $x,y,p,q>1$ in his paper

  • R. Tijdeman, On the equation of Catalan, Acta Arith. 29 (1976) pp 197–209 (EuDML)

He just found the following upper bound for $p$ and $q$ using Baker theorem in linear form in logarithm

\begin{align} p& <2c_{9}(\log p)^{c_{10}}\\ q& <c_{1}(\log p)^{c_{2}} \end{align}

I don't understand how these two inequalities for $p$ and $q$ give us that Catalan's equation has only a finite number of solution since he didn't give an upper bound for $x,y$.

Also in 1993, Overholt showed that Brocard equation $$ n!+1=m^{2} $$ has finitely solution if Szpiro's conjecture is true. He just found that $n<4^{\epsilon}e $. I don't understand how this upper bound for $n$ make us say that the Brocard equation has finitely many solutions?

I ask if a finding of an upper bound for a least one variable of an arbitrary Diophantine equation is enough to prove that it has only finitely many solutions in $\mathbb{Z}$? If yes does the upper should do not depend on the other variable of that Diophantine equation?

Edit I need to answer the third question just .

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Faltings%27s_theorem $\endgroup$ – YCor Aug 22 at 18:48
  • $\begingroup$ Please restrict to one question per post (standard MO policy). I answered your first question below. $\endgroup$ – GH from MO Aug 22 at 19:26
  • $\begingroup$ @Ycor which theorem do you mean $\endgroup$ – Abdo Aug 22 at 20:55
  • 1
    $\begingroup$ For the Brocard equation, for each $n$, how many values of $m$ are possible? Answer: 2. Hence if you bound $n$, then you know there are finitely many solutions. If you want an explicit bound, just use $m=\sqrt{n!+1}<\sqrt{4^\epsilon e+1}$ (assuming Szpiro). $\endgroup$ – Joe Silverman Aug 22 at 21:42
  • 1
    $\begingroup$ If a Diophantine equation has two variables $x$ and $y$, and you find an upper bound for $y$, then in effect you reduced the original equation to an equation in $x$. Now a one-variable Diophantine equation has finitely many solutions (unless it is degenerate like $x=x$), so the answer to your last question is "yes". $\endgroup$ – GH from MO Aug 22 at 22:54

By the initial remarks in the paper, one can restrict to $p,q\geq 5$. By Theorem A in the paper (which is a result of Baker's from 1969), $x$ and $y$ can be effectively bounded in terms of $p$ and $q$: $$\max(|x|,|y|)<\exp\exp(5^{10}p^{10} q^{10q^3}).$$

Hence it suffices to bound $p$ and $q$.

| cite | improve this answer | |
  • $\begingroup$ but i mean in general is it enough to find an upper bound for only one varible (3rd question ) $\endgroup$ – Abdo Aug 22 at 20:54
  • 2
    $\begingroup$ @Abdo Yes, in some sense, but given the form of the Brocard equation, once you give an upper bound for one of the variables, it's trivial to get an upper bound for the other one. Similarly, on an elliptic curve, Siegel's theorem say $y^2=x^3+Ax+B$ has only finite many solutions $(x,y)$ in integers, but effective versions often just write down a bound for $|x|$. However, it's possible sometimes to prove finiteness of solutions without bounding the size of the solutions. Indeed, Sigeel's original proof of his theorem has this form. The trick: Show $E(\mathbb Z)$ is a subset of a finite set ... $\endgroup$ – Joe Silverman Aug 22 at 22:47
  • 1
    $\begingroup$ ... whose number of elements can be bounded, but for which you might not be able to find an upper bound for the size of the largest element. This is typical of proofs in Diophantine approximation, which is often used to study solutions to Diophantine equations. $\endgroup$ – Joe Silverman Aug 22 at 22:48
  • 1
    $\begingroup$ @Abdo For an equation such as $n!+1=m^2$, an upper bound for one variable trivially gives an upper bound for the other variable. Similarly for an equation such as $x^3-2y^3=107$. But the original proof that $x^3-2y^3=107$ has finitely many solutions did not actually give an upper bound for $x$ and $y$. ... $\endgroup$ – Joe Silverman Aug 23 at 19:19
  • 1
    $\begingroup$ ... Instead, it proved (roughly) that there were effectively computable constants $a$ and $b$ such *if there were a solution with $\max\{|x_0|,|y)_0|\}\ge a$, then every solution satisfies $\max\{|x|,|y|\}\le b\cdot\max\{|x_0|,|y)_0|\}$. This obviously shows that there are finitely many solutions, without given an explicit upper bound for the largest $\endgroup$ – Joe Silverman Aug 23 at 19:20

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.