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May you tell me if my answer is correct? Thank you so much!

Here is the problem:

Given f(x)=

enter image description here

where the constant term is the product of r distinct primes,

determine the minimum degree of f(x)

such that is guaranteed to have at least one irrational root.

Here is my solution:

By the Rational Root Theorem:

Rational Root Theorem

As the leading coefficient is 1, and the constant term is the product of r distinct primes.

We can use this two theorems about divisors and the tau function:

Divisors

Tau function definition

enter image description here

Thus to guaranteed to have at least one irrational root, we need a polynomial of degree (2 times the tau function) + 1

Note: two times because we need to consider positive and negative numbers.

Beginner
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    Also, $105 = 3 * 5 * 7$ has 8 positive integer factors and $8$ negative integer factors. Your formula would predict only 10. – Ken Duna Apr 27 '17 at 18:48
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    I assume you also want $a_i\in \mathbb Z$? – lulu Apr 27 '17 at 18:51
  • @KenDuna Thanks! You are mentioning a very important mistake... – Beginner Apr 27 '17 at 18:51
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    But what's to stop me from taking something like $(x-1)^{100}(x+2)$? That has no irrational roots and the constant term is $2$. – lulu Apr 27 '17 at 18:54
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    Unless you block multiply roots I do not believe there is an upper bound. If you do block multiple roots I think $r+2$ will do the job. – lulu Apr 27 '17 at 18:56
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    As I said, the claim is false as it stands. $f(x)=(x-1)^k$ has constant term of your form and $r=0$, yet it can have arbitrarily high degree and no irrational roots. – lulu Apr 27 '17 at 19:14
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    The easiest way I can see to get to a true bound is to add the assumption that $f(\pm1)\neq 0$. Now it works, and $r+1$ is an upper bound. – lulu Apr 27 '17 at 19:15
  • @lulu Your solution is a counterexample! A powerful one... Then the solution is that there is no solution unless f(1) and f(-1) are not equal to 0. If that condition is satisfied, the degree would be 2*(the tau function less 1) + 1. What do you think? Thanks! – Beginner Apr 27 '17 at 19:39
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    Still no. If you add my assumption then $r+1$ is the minimum (for square free $a_0$). The tau function is not relevant, nor are the signs. If you take the case in which $a_0$ is the product of $r$ distinct primes then you can't have any two roots divisible by the same prime (lest the square of that prime divide the constant term). Similarly in the general case, the bound is much lower than tau. – lulu Apr 27 '17 at 19:43
  • @lulu The problem is why a0 is square free? The tau function covers all the possible cases. – Beginner Apr 27 '17 at 19:45
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    Yes, but it covers them incorrectly. For $a_0=6$ we'd have $\tau (6)=4$ but you can't have $2,3$ and $6$ as roots (as the constant term would then be divisible by $6^2$. – lulu Apr 27 '17 at 19:46
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    In general, if $a_0=\prod p_i^{b_i}$ then the minimum is $1+ \sum b_i$, or something close to that. – lulu Apr 27 '17 at 19:47
  • @lulu What if we use the tau function less 2 to eliminate 1 and a0? Thanks! – Beginner Apr 27 '17 at 19:52
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    No, really. I think I have it right with my expression. with $a_0=\prod p_i^{b_i}$ and excluding $\pm 1 $ as roots the highest degree you can get is with $f(x)=\prod (x-p_i)^{b_i}$ and that has degree $\sum b_i$, so adding $1$ to that implies an irrational root. – lulu Apr 27 '17 at 19:55
  • @lulu Thank you for your amazing help! – Beginner Apr 27 '17 at 20:02

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