Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

consider some function like $y = x^2 + 3x$

and then some family of related polynomial functions

(like: $y = x^2 + 3x$, $y = 2x^2 + 4x$, $y = 3x^2 + 5x$, $y = 4x^2 + 6x$, etc.)

what method or set of tools would conventionally be used for determining/defining all integers $y$ that CANNOT be produced by any integer $x$ in any one of these functions?

I understand that the study of Diophantine equations seems sort of relevant, but it also seems like it focuses more on the opposite (the solutions, not the $y$'s that can't be solutions), and in any event I don't understand what specific methods it would offer to tackle this kind of problem.

share|improve this question
    
Any thoughts on the answers you've received in the last few days? –  Gerry Myerson Jul 9 '12 at 7:27
add comment

3 Answers

The answers to can and cannot are so closely related that it is difficult to separate them. Let us deal with $x^2+3x$. Other quadratics yield to similar methods.

There is a simple answer to what numbers can be represented in the form $x^2+3x$, where $x$ is an integer. Rewrite the equation $y=x^2+3x$ as $4y=4x^2+12x$, and note that $4x^2+12x=(2x+3)^2-9$. So our equation becomes $$(2x+3)^2=4y+9.$$ Thus to determine whether $y$ is of the right shape, all we need to do is to check whether $4y+9$ is a perfect square. For if $4y+9$ is a perfect square, it will be an odd perfect square, and therefore there will be an integer $x$ such that $(2x+3)^2=4y+9$.

So one answer to what numbers $y$ cannot be represented as $x^2+3x$ is: the $y$'s such that $4y+9$ is not a perfect square. It is not a fully satisfactory answer, since it is in some ways too close to the original question.

Remark: We can reach the same conclusion by rewriting our equation as $x^2+3x-y=0$. then solve using the Quadratic Formula. We get $$x=\frac{-3\pm\sqrt{9+4y}}{2}.$$

share|improve this answer
add comment

One thing that you could try is to find all integers that are expressible in one of the above forms. THen, after you do this, you'll have your set by taking the complement.

share|improve this answer
add comment

I don't think there is any method, or even any finite collection of methods, that will work for all families of "related" polynomials, but here's a way to do the specific family mentioned in the question.

First, note that you can't get any odd numbers at all; each polynomial is either twice an integer-coefficient polynomial, or $x^2+x$ plus twice an integer-coefficient polynomial, and $x^2+x=x(x+1)$ takes on only even values. So, now it's just a question of what even numbers we get.

Next, note that $x=1$ gives you $4,6,8,10,\dots$, that is, all the even numbers, starting with 4.

$x=0$ gives you 0, and $x=-1$ gives you $-2$. You get 2 by evaluating $3x^2+5x$ at $x=-2$. So now we have all the even numbers starting with $-2$.

Then it's not hard to show that these polynomials don't give any numbers less than $-2$, so we're done.

Very ad hoc, I know, but presumably adaptable to other families of polynomials.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.