Quadratic equation with no rational roots

[brotherbobby]

Homework Statement:

Prove that the equation $x^2+(2m+1)x+(2n+1) = 0$ does not possess any rational roots if $m \in \mathbb{Z}, n \in \mathbb{Z}$.

Relevant Equations:

1. Discriminant of the quadratic equation $ax^2+bx+c = 0$ is $\mathscr{D} = b^2-4ac$
2. If the roots of a quadratic equation are rational, its discriminant must be a perfect square (of an integer), i.e.$\mathscr {D} = p^2$ where $p \in \mathbb{Z}$.

Given : Equation $x^2+(2m+1)x+(2n+1) = 0$ where $m \in \mathbb{Z}, n \in \mathbb{Z}$, i.e. both $m,n$ are integers.
To prove : If $\alpha,\beta$ be its two roots, then they are not rational numbers.

Attempt : The discriminant of the equation $\mathscr{D} = (2m+1)^2 - 4(2n+1) = 4m^2+4m+1 - 8n - 4 = 4m^2+4m-(8n+3)$. I need to show therefore that $4m^2+4m-(8n+3) \ne p^2$, where $p\in \mathbb{Z}$.

If possible, let $4m^2+4m-(8n+3) = p^2$. Being a quadratic equation with an integral root ($m\in \mathbb{Z}$), the solution of this equation $m = \frac{-4 \pm \sqrt{4^2+16(8n+3+p^2)}}{8} = \frac{-1\pm \sqrt{4+8n+p^2}}{2}$ which implies that the factor within the square root $4+8n+p^2$ must be an odd integer squared, i.e. $4+8n+p^2 = (2r+1)^2$ where $r \in \mathbb{Z}$.

Rearranging, we get $p^2 = 4r^2+4r-8n-3$. Is the expression on the right a perfect square, for the sake of p?

I find myself stuck here. A help would be welcome.

 

[PeroK]

Didn't we prove recently that if such an equation has rational roots, the roots must be integers?

Note: $2m +1, 2n + 1$ are both odd integers.

 

[brotherbobby]

Didn't we prove recently that if such an equation has rational roots, the roots must be integers?

Note: 2m+1,2n+1 are both odd integers.

Yes we did. I copy and paste from my earlier post.

I can't relate the two problems. Are they converse of one another? Either way, the current problem still needs to be proved.

 

[PeroK]

I can't relate the two problems.

Really? They are just two random problems with nothing in common?

 

[brotherbobby]

Really? They are just two random problems with nothing in common?

Ok let me see closely. I will start with the earlier problem first.

1. Given a quadratic equation with integral coefficients and told that it has rational roots. Then we show that the roots must be integers also.

2. Given an equation with (odd) integral coefficients. Nothing is given about the roots as to whether they are rational or not. We cannot conclude that the roots are therefore integral. In fact, we are asked to prove that the roots are not rational.

I cannot see how information or ideas from the first would help me solve the second problem. Thank you for your time and patience.

 

[brotherbobby]

Haha. Sir @PeroK , please forgive me. I don't see the way out. Do you think I should spend a few hours on it thinking and then return to you?

 

[PeroK]

This is what makes mathematics difficult. It is genuinely abstract and requires a way of thinking that is unnatural to many people (the vast majority of people?). Ultimately, there's only so much I can do. If we think about this rationally, eventually your neurons have to be forced to align themselves in a certain way and there's no telling how long that will take!

It's a serious point: if you don't see it, you don't see it.

There's no problem with me explaining this:

1) If the equation $x^2 + (2m + 1)x + 2n + 1 = 0$ has a rational solution, then that solution must be an integer.

2) Let $r \in \mathbb Z$ be a solution. And, we have: $r^2 + (2m + 1)r + 2n +1 = 0$ and, rearranging we have $r(r + 2m + 1) = -(2n +1)$

3) Note that $-(2n+1)$ is odd.

4) If $r$ is odd then ... contradiction. I'll leave this bit to you.

5) If $r$ is even ... contradiction.

6) The equation cannot have an integer solution, therefore cannot have a rational solution, therefore any solution is irrational.

That's what's generally called an example of mathematical logic. Although it feels very natural to me, it is perhaps quite abstract and esoteric to a lot of people!

 

[haruspex]

Really? They are just two random problems with nothing in common?

Well, no, but doesn't applying the earlier result only get as far this point already reached in post #1:

I need to show therefore that $4m^2+4m-(8n+3) \ne p^2$, where $p\in \mathbb{Z}$.

?
@brotherbobby , what can you say about perfect squares modulo 8?

 

[PeroK]

Well, no, but doesn't applying the earlier result only get as far this point already reached in post #1:

An important aspect of pure mathemetics is to build on previous results and not start every problem from first principles. I'm pretty sure whoever is setting these problems expected the OP to use the earlier result here. I think it is really important for the OP to start thinking like this.

Also, in the previous problem there was the same issue of the OP diving headlong into complicated algebra, rather than using the properties of numbers.

If the OP is studying number theory, then it's important to develop a pure-mathematical (rather then a mathematical-methods) mode of thinking.

 

[brotherbobby]

Well, no, but doesn't applying the earlier result only get as far this point already reached in post #1:

?
@brotherbobby , what can you say about perfect squares modulo 8?

I am afraid I don't know. Sorry for coming in late for the discussion. I am the creater of this thread. What's the meaning of OP?

I am aware of this though - when we say that "a is congruent to b mod n" ($a \equiv b \mod n$) we imply that "n divides a-b" ($n | (a-b) \Rightarrow a-b = kn, k \in \mathbb{Z}$).

You asked for the perfect squares modulo - I haven't heard of that.

 

[brotherbobby]

An important aspect of pure mathemetics is to build on previous results and not start every problem from first principles. I'm pretty sure whoever is setting these problems expected the OP to use the earlier result here. I think it is really important for the OP to start thinking like this.

Also, in the previous problem there was the same issue of the OP diving headlong into complicated algebra, rather than using the properties of numbers.

If the OP is studying number theory, then it's important to develop a pure-mathematical (rather then a mathematical-methods) mode of thinking.

Dear @PeroK , sorry for coming in late. Thank you for working out the solution in Post#7 above. I am afraid I would not have been able to do it myself.

Please tell me what OP stands for. No, I am not studying number theory - that is advanced math. These are school level problems from a russian textbook. I must admit they are higher than school level, but they can be solved by a student using his school mathematics in a thorough way. I agree with you that the author is seeking to build on results in subsequent problems. Please bear with me as mathematics is not my major.

 

[PeroK]

Dear @PeroK , sorry for coming in late. Thank you for working out the solution in Post#7 above. I am afraid I would not have been able to do it myself.

Please tell me what OP stands for. No, I am not studying number theory - that is advanced math. These are school level problems from a russian textbook. I must admit they are higher than school level, but they can be solved by a student using his school mathematics in a thorough way. I agree with you that the author is seeking to build on results in subsequent problems. Please bear with me as mathematics is not my major.

OP is the orginal poster and in this case that's you!

It may be advanced maths, but it's number theory once you start talking about rational and integer solutions. In any case, it's pure mathematics, rather than applied maths. In applied maths, you're generally not concerned about integer solutions, as opposed to real, irrational solutions. [I'm sure there's a counterexample to this, but it's true in most cases.]

Fundamentally, this sort of pure mathematics is a different ballgame from applied maths.

 

[haruspex]

I am afraid I don't know. Sorry for coming in late for the discussion. I am the creater of this thread. What's the meaning of OP?

I am aware of this though - when we say that "a is congruent to b mod n" ($a \equiv b \mod n$) we imply that "n divides a-b" ($n | (a-b) \Rightarrow a-b = kn, k \in \mathbb{Z}$).

You asked for the perfect squares modulo - I haven't heard of that.

"Modulo" gets used in two slightly different ways.
The number "m modulo n" is that number for 0 to n-1 which is congruent to m modulo n.
So do a little experimentation. Run through the perfect squares of 1 to 7 and see what they are modulo 8. Can you see why the pattern must repeat after that, and why only those values occur?