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There are at most 3 perfect squares in arithmetic progression (Fermat, Euler). It was shown in [1] that if $n>2$ there are no three term arithmetic progression consisting of nth powers.

Take a non-linear polynomial $p(x)$ with integer coefficients and consider the polynomial set $ A_p = p(\mathbb{N}) $. Let $f(A)$ be the maximal length over all arithmetic progressions whose elements are in $A$. It is not hard to prove, reducing it to the perfect squares case, that if the $\deg(p)=2$ then $f(A_p)<4$.

Question 1: Is it true that $f(A_p)$ is always finite if $\deg(p) >1$ ?

It is clear that there are no infinite arithmetic progressions inside polynomial sets, but I do not see how to exclude the possibility of having arbitrarily large AP.

Question 2: Is there a uniform bound for $f(A_p)$ over all polynomials $p(x)$ with $\deg(p)=k$ ?

For example, can we find $10$ terms in arithmetic progression whose elements belong to some polynomial set $A_p$ with $\deg(p)=3$?

[1] H. Darmon and L. Merel, Winding quotients and some variants of Fermat’s Last Theorem, J. Reine Angew. Math. 490 (1997), 81–100.

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If there was no uniform bound for $f(A_p)$ for degree three polynomials, then (after scaling etc) I could find a cubic $f(x)$ and rationals $x_1$, $x_2$, $x_3$...$x_{1000000}$ such that $f(x_n)=n$. Now because 1000 of these $n$ are squares I have now got 1000 points on $y^2=f(x)$ and now perhaps I am pretty close to constructing elliptic curves with arbitrarily large rank. It's unclear to me what the concensus is about these things existing. Similarly I can construct curves of genus 3, say, over Q with as many points as you like and again it's not clear that such things should exist. –  Kevin Buzzard Mar 24 '11 at 21:50
    
Don't terms in arithmetic progression lie on a straight line, and we are talking about the intersection of a polynomial with a straight line? –  Mark Bennet Mar 25 '11 at 19:08
    
This doesn't help with your problem, but it is interesting to note that as soon as we allow two variables, we get arbitrarily long arithmetic progressions from only a quadratic example: the polynomial $x^2+y^2$, for example. This is an immediate consequence of the Green-Tao theorem. –  Thomas Bloom Mar 25 '11 at 20:02
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@Mark: The points do not have to be on a straight line. Take for example (1,1), (5,25) and (7, 49) which correspond to the 3-AP of squares 1, 25, 49. –  Manuel Silva Mar 25 '11 at 20:05
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4 Answers

[Edited to give direct connection with Caporaso-Harris-Mazur via Kevin Buzzard's idea]

Such results are probably true but out of reach of present-day techniques. For each $p$ or $k$, and every sufficiently large $n$ (say $n>k$), the $n$-term arithmetic progressions of the desired type are parametrized by nontrivial points on some algebraic variety, call it $V_n$, of fixed dimension: dimension $2$ if $p$ is fixed (assuming it's not of the form $a(x-x_0)^k+b$ in which case we're back to Darmon-Merel), and degree about $k$ if $p$ is allowed to vary over all polynomials of degree $k$. In each case we have for each $n$ two maps $V_n \rightarrow V_{n-1}$ of degree $k$ that forget the first or last term of the progression; and $V_n$ should be of general type for $n$ large enough. We're now in a setting similar to that of this recent Mathoverflow question (#73346), and I give much the same answer as I did for that question: the claim should follow from the Bombieri-Lang conjectures plus some possibly nontrivial extra work, as in

L.Caporaso, J.Harris, and B.Mazur: Uniformity of rational points, J. Amer. Math. Soc. 10 #1 (1997), 1-45

but (excluding some very special cases that don't seem relevant here) we have no techniques for proving such results unconditionally on varieties of dimension greater than 1.

EDIT Indeed this is a special case of Caporaso-Harris-Mazur, by adapting Kevin Buzzard's observation in his comment on the original question: write the equations as $f(x_m)=m$ $(m=1,2,\ldots,n)$ for some degree-$k$ polynomial $f$ (obtained from $p$ by suitable translation and scaling), and consider just those $m$ in that range which are of the form (say) $y^5$ or $y^5+1$. By Mason's theorem (polynomial ABC) either $f$ or $f-1$ has at least two zeros whose order is not a multiple of $5$, so either $f(x) = y^5$ or $f(x)=y^5+1$ defines a curve of genus at least $2$. But the genus is clearly $O(k)$ for any polynomial $f$ of degree $k$. So, if the number of rational points on such a curve has a uniform bound, then so does the length of an arithmetic progression of values of a polynomial of bounded degree.

Come to think of it, the reduction to C-H-M could also be obtained more directly from the curve $p(x')-p(x)=d$ (for $k>3$), or $p(x''')-p(x'')=p(x'')-p(x')=p(x')-p(x)=d$ (to cover $k=2$ and $k=3$ as well), where $d$ is some multiple of the common difference of the arithmetic progression.

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Doesn't Lagrange interpolation tell you that you can fit $n+1$ numbers (even if $n$ are in an arithmetic progression) to a non-linear polynomial of degree $n$? and if the common difference is divisible by the right numbers, said polynomial will have integer coefficients?

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An edit made this a bit unclear: The original form of the question makes clear that question 2 is meant as unform for for a fixed k. –  quid Mar 24 '11 at 23:45
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@Gerry: The polynomial degree is fixed. I have changed the original formulation to make it more clear. –  Manuel Silva Mar 25 '11 at 0:51
    
Oops.${}{}{}{}$ –  Gerry Myerson Mar 25 '11 at 5:47
    
Polynomial of degree at most $n$ - you will get a straight line surely? –  Mark Bennet Mar 25 '11 at 19:09
    
@Mark, note it's $n$ numbers in AP, together with one more number not in the AP, so the polynomial of degree (at most) $n$ won't be a straight line. –  Gerry Myerson Mar 25 '11 at 21:48
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For a very particular case, when de degree of the polynomial is 3 and the arithmatic progression has reason not congruent to 3, then the polynomial can't have an arithmetic progression with more than 4 elements. I can send you the proof if you want it.

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Post please! . –  Kevin O'Bryant Oct 2 '11 at 1:24
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I think I have a proof, given by a friend of mine. I’m going to post the proof when the degree is 3, but it is easy to do a generalization.

Assume $p(x)=ax^3+bx^2+cx$, and for all $i\in\{0,1,2,3\}$ we have $p(a_i)=ir$. Consider the equation $$p(x)-ir=0 \iff bx^2 + cx + (ax^3-ir)=0 \iff x=\frac1{2b}\left(-c \pm \sqrt{c^2-4b(ax^3-ir)}\right),$$ if we calculate the derivative in order to $x$ we have: $$1=\frac1{4b}\frac{-12b\cdot ax^2}{\sqrt{c^2-4b(ax^3-ir)}} \iff \sqrt{c^2-4b(ax^3-ir)}=-3ax^2,$$ so in the equation $x=\frac1{2b}\left(-c \pm\sqrt{c^2-4b(ax^3-ir)}\right)$ we have $x=\frac1{2b}(-3ax^2)$.

Every one of the $a_i$’s not equal to zero must satisfy the last equation, since it is a degree $2$ polynomial we only have, at maximum, $2$ solutions, hence a arithmetic progression must have maximal length equal to $3$.

The general case follows more or less in this way, we use the $2$-degree formula, and take the derivative in order to $x$, which will give us a polynomial equation of $n-1$ degree.

Sorry, but I can't write LATEX... Please tell me if it is correct.

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The reasoning is not correct: the equation $p(x)-ir=0$ only holds for the isolated points $a_i$, not in their neighbourhoods, hence taking derivatives will not preserve the equality. Note that your argument never used the assumption that $a_i$’s are integers, but then the result does not hold, there are infinite arithmetic progressions with real $a_i$ (since the range of $p$ is all of $\mathbb R$ or a one-side unbounded interval). –  Emil Jeřábek Jul 13 '12 at 12:56
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