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I have the following problem from USAMO 2006:

"For an integer $m$, let $p(m)$ be the greatest prime divisor of $m$. By convention, we set $p(\pm 1)=1$ and $p(0)=\infty$. Find all polynomials $f$ with integer coefficients such that the sequence $\{ p(f(n^2))-2n) \}_{n\ge 0}$ is bounded above. (In particular, this requires $f(n^2)\neq 0$ for $n\ge 0$.)"

One solution for the above question is given here:

https://artofproblemsolving.com/wiki/index.php?title=2006_USAMO_Problems/Problem_3"

In this solution, the author makes the following statement:

"A theorem of T. Nagell implies that...." and has continued proving the given statement. Can anybody help me with what theorem of Nagel being referred to here and how it is used in the above solution?

PS: The only place where I have encountered the name before is the Nagell point, which for sure is not at all related to this question.

the_fox
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saisanjeev
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1 Answers1

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The theorem of T. Nagell referred to is:

Theorem (T. Nagell, Généralisation d'un théorème de Tchebycheff, J. Math. Pures Appl. (8) 4 (1921), 343-356) Let $f$ be an irreducible integral nonlinear polynomial, and let $P_x$ be the greatest prime factor of $$ \prod_{n\leq x}f(n). $$ Then, for any $a<1$ and all sufficiently large $x$, $$ \frac{P_x}{x}>(\log x)^a. $$

This bound is further improved upon by Erdős and others later but we don't need that here. You should see immediately how Nagell's result is applied in the solution.

And no, Nagel (the 19th century German geometer Christian Heinrich von Nagel) $\neq$ Nagell (Trygve Nagell, the 20th century Norwegian number theorist).

user10354138
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  • Thanks for the reference. I want to know the proof for this, but is it too high a level for a high school student? – saisanjeev Oct 02 '18 at 08:00
  • Can you also help me with the following statement in the above proof: " since $P(0) = \infty$, neither polynomial can have any non-negative integer roots, so $a_i > 1$ and thus $b_i > 0$". How has he concluded the last part? – saisanjeev Oct 02 '18 at 08:09
  • For quadratic $f$, you can proceed similar to Chebyshev's $f(n)=n^2+1$ case (I'm assuming you are familiar with the elementary analytic number theory as covered in, e.g., Hardy&Wright). For higher degree I don't think I can safely say it is accessible to secondary school student. – user10354138 Oct 02 '18 at 08:35
  • For your second question, recall $a_i\geq 1$ (otherwise you could change the sign of both $g_i$ and $h_i$), so if $a_i=1$ there would be roots $\pm b_i$ of $g$ and one of them would be nonnegative. So $a_i>1$. Also recall $b_i\geq 0$ is assumed, so if $b_i=0$ we would have $g_i(0)=0$ which is not allowed. Thus $b_i>0$. – user10354138 Oct 02 '18 at 08:43