0
$\begingroup$

Let us define a power Diophantine equation by 2 algebraic functions $f,g$ (having different degree) and by integers $k, l >0$ where, there are finite solutions for $f(x)+k=g(y)$, but there exists $l$ such that- there are infinite integer solutions for the equation $f(x)+l=g(y)$. One may add more functions of different variable, and alter sign in the equation.

Do such power Diophantine equations exist? What are examples of such power Diophantine equations?

For example, Tijdeman's theorem states that there are at most a finite number of consecutive powers. If we write that, $x^m=y^n+1,$ then $f(x)=x^m, g(y)=y^n, k=1$, though it is not known, for $k>1$. So this doesn't fit the Diophantine Equations I am looking for.

If the post requires more specification, please let me know.

Improve this question
$\endgroup$
8
  • $\begingroup$ I assume you mean integer solutions. Then $x^2-1=2y^2$ has infinitely many solutions and $x^2-2=2y^2$ has finitely many (actually no) solutions. The equation $x^2+k=-y^2$ has finitely many solutions for $k\ne0$ and infinitely many solutions for $k=0$. $\endgroup$
    – Joe Silverman
    Commented Nov 17, 2019 at 20:09
  • 1
    $\begingroup$ @Joe, I don't think $x^2+k=-y^2$ has infinitely many solutions for $k=0$. $\endgroup$
    – Gerry Myerson
    Commented Nov 17, 2019 at 21:16
  • 1
    $\begingroup$ I don't know what you mean by "informing", Jim, and I don't see anything in your question indicating what does and what doesn't count as an answer to your question. $\endgroup$
    – Gerry Myerson
    Commented Nov 17, 2019 at 21:18
  • 1
    $\begingroup$ @GerryMyerson Right. Messed up the sign. $x^2+k=y^2$ has finitely many solutions for $k\ne0$, since $(x-y)(x+y)=-k$ yields two equations for $(x,y)$, and of course has infinitely many solutions $(t,t)$ for $k=0$. $\endgroup$
    – Joe Silverman
    Commented Nov 17, 2019 at 21:25
  • 1
    $\begingroup$ Italicizing is used to emphasize. Parentheses are used to put aside something less important or optional. Using italics exactly for a parenthesized part thus looks funny. Also "algebraic function" seems to be used for "polynomial function" (in one variable). $\endgroup$
    – YCor
    Commented Nov 17, 2019 at 22:09

1 Answer 1

Reset to default
1
$\begingroup$

I'll assume that by "algebraic functions" is meant polynomials with integer coefficients. If both $f$ and $g$ have degree at least two, and at least one of them has degree at least three, then, trivial cases aside, it's well-known that the equation has only finitely many integer solutions.

If both polynomials have degree two, then there is elementary theory available to decide whether there are finitely many or infinitely many solutions – this can better be discussed elsewhere, e.g., math.stackexchange.com (where it probably has already been discussed, repeatedly).

If both polynomials have degree at most two, and at least one has degree less than two, then it's even easier to decide the number of solutions, and again math.stackexchange.com would be the place to ask (or to look to see whether there's an answer there already).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks, can you refer me to related paper/book/theorem (keywords/ author name to search) , specially for the first part? Also, if you can improve the headline of the post for future users, please do that. $\endgroup$
    – Michael
    Commented Nov 18, 2019 at 17:50
  • $\begingroup$ Siegel's Theorem, start at en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points $\endgroup$
    – Gerry Myerson
    Commented Nov 18, 2019 at 22:33

You must log in to answer this question.

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