Background
This is an experiment where I narrate my thought process as I solve a math problem. My goal: writeups of the solutions to math problems usually present a polished, streamlined version of the solver's thought process that omits errors, wrong turns, and heuristics. I wondered what it would be like to represent the thought process "warts and all."
In full disclosure, I didn't get the idea to liveblog my solving of this problem until after the "IS THAT IT??" insight, so the beginning is my attempt to reconstruct what I was thinking then.
I encountered today's problem here.
Problem
Prove that if $a$, $b$, and $c$ are odd integers, then the roots of $ax^2 + bx + c = 0$ are irrational.

Interesting; so this means that the graph of $ax^2 + bx + c$ never intersects the xaxis at an integer point.
 Not sure how to proceed from that; maybe we need to do something algebraic.

Oh, obviously we can use the quadratic formula. $\frac{b^2  4ac}{2a}$. No, $\frac{b \pm \sqrt{b^2  4ac}}{2a}$. The part that matters is inside the square root. We just need $b^2  4ac$ not to be a perfect square; if it isn't a perfect square, then neither will the roots be.

Okay; we have that $a, b, c$ are odd integers. They can be negative or positive, I notice.

Well, an odd integer squared is still odd. So $b^2$ is odd. And $4ac$ is even, since it's an even number times an odd number times an odd number.

Is that right? Yes, it is, because evenness spreads like poision during multiplication (remember this via $1 \times 0 = 0$, odd times even is even).

Okay, so $b^2  4ac$ is an odd minus an even, so it's odd.

Wait; this doesn't get us anywhere. I don't care whether it's odd or not; I care whether it's a square.


Unsure what to do.

Okay, let's try with some actual numbers. I need two squares and I know one of them has to be odd; let's use 9 for our odd square. I'll take another square — say 25 — and try to subtract enough to get it to 9.

What do I have to subtract? Right; 16.

Okay; that won't work. I have to get $4ac$ to equal 16, but both $a$ and $c$ are odd whereas I need them both to be 2 in order to get 16.


Is there some sort of argument about how large the numbers can get? Like, if it's below 100 I can't hit it in this way, if it's above I can't hit it in another way? Some sort of bipartite argument.
 No, that's tough because there are two things that vary; the large square which I'm subtracting $4ac$ from and the small square I'm trying to hit.

Let's choose another example. What's the next biggest square after 25? 36. Okay, I can't subtract from 36 to get to 25 because I'd have to subtract 11, and that's too big to be hit by $4ac$, I've already seen that.

Can I subtract to get to 9? No, I'd have to subtract 27, and that doesn't work either.

Wait, there's another reason why these don't work; the amounts I'd have to subtract are all odd, but I need them to equal $4ac$.

IS THAT IT??


Right; when you list squares they look like this: 0, 1, 4, 9, 16 ... and the deltas look like this: 1, 3, 5, 7 ...

It's all the odd numbers.

Imagine a sentence in the writtenup proof. "Consider the set of differences of squares..." Is it generally the case that the difference of any two squares is an odd number? Not just counting consecutive ones.

I don't think so; for example $49  9 = 40$ is not odd.


Perhaps we have to use the fact that $b^2$ is odd. But it's troubling that $40$ is not odd in the example above.
 Why doesn't the example above work? That is, why can't we have $b^2 = 49$ and $4ac = 40$? Well, $a$ would need to be $5$ and $c$ would need to be $2$ for that to work, but $c$ can't be $2$. So we're saved, but how are we saved in general?

Okay, let's dive into the algebra. $b$ is odd, so say it's $2k + 1$.

Then $b^2 = (2k + 1)^2 = 4k^2 + 4k + 1$.

I guess we'll subtract a square out and show that the result is not even? Or subtract an even number out and show that the result is not a square?

First one seems easier. But wait; how to subtract a square out?

Okay, let's try the second one.

Hmm, how do I subtract an even number out? I feel like that trailing 1 is giving me what I want. But... $4k^2 + 4k$ will be an even + an even = an even, so if I leave that out I get 1, which is a square.
 Why doesn't this present a counterexample? Perhaps we simply can't find $a, c$ such that $4ac = 4k^2 + 4k$, but I am at a loss as to why.


Feel like I've taken a bad turn somewhere. Let's lay out what we've got in algebra. Also, let me head into another room.

We need to show that $b^2  4ac \ne d^2$, where $a, b, c$ are odd integers and $d$ is any integer.

We can rewrite this into $b^2  d^2 = 4ac$ and trying to show a contradiction, which is the direction we were going above. "It would be absurb to take an odd square, subtract a square from it, and get $4ac$" (the product of two odd integers and 4).

What can we say about the differences of squares? Are they ever just 4 (so $a = c = 1$)? Well, no, as I laid out above there are no consecutive squares distance 4 apart, and the nonconsecutive ones are too far apart to be 4 apart.

Can we just do odd and even? We know $b^2$ is odd, so subtract another square from it. If we subtract an even square, the result is odd and can't be $4ac$. If we subtract an odd square, things get more interesting...

Can we guarantee anything about relative size of $b$ and $d$? I don't think so...


Subtract an odd square from an odd square. Say, $4k^2 + 4k + 1  (4j^2 + 4j + 1)$. The $1$s go away and we can take out a 4, so $4(k^2 + k  j^2  j)$. We want to show this can't be $4ac$?
 Well, $k^2 + k$ is even regardless of what $k$ is, since it's odd plus odd. And same goes for $j^2 + j$. So it's 4 times an even minus an even = an even; a poor fit for $4ac$, I suppose.

This seems like it. But how to we justify that this cannot be $4ac$?
 Oh, we can simply divide by $4$. So, when you divide $4ac$ by 4 the result is odd; when you divide $4(k^2 + k  j^2  j)$ by 4 we know that the result is even. Hence they are different.

(Took a break)

Great, I think we can write up the final solution now.
Solution
By the quadratic formula, we just need to prove that $b^2  4ac$ is not a perfect square. We proceed by contradiction. Suppose it were; that is, suppose $b^2  4ac = d^2$, where $d$ is any integer. Rearrange: $b^2  d^2 = 4ac$. There are two cases to consider: if $d$ is even, and if $d$ is odd.

If $d$ is even, then $d^2$ is even. But then the lefthand side is odd (odd  even) and the right side is even, so there is a contradiction.

If $d$ is odd, then write $2k + 1$ for $b$ and $2j + 1$ for $d$, where $j, k$ are any integer. Expanding, we get
$$b^2  d^2 \rightarrow 4k^2 + 4k + 1  (4j^2 + 4j + 1) \rightarrow 4(k^2 + k  j^2  j) = 4ac$$

Divide both sides by 4; $k^2 + k  j^2  j = ac$

Now, observe that the quantity $k^2 + k$ must be even, since $k^2$ has the same parity as $k$. Same with $j^2  j$. Therefore lefthand side $k^2 + k  j^2  j$ is even.

But the righthand side $ac$ is an odd number multiplied by an odd number and is therefore odd. Hence, a contradiction.
Learnings
The difference of two odd squares is always a multiple of four; that's pretty interesting. (Furthermore, it's four times an even number, so I guess it's a multiple of eight.) I suppose the difference of two even squares is a multiple of four as well.