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Math Problem Liveblogging #1

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 aa, bb, and cc are odd integers, then the roots of ax2+bx+c=0ax^2 + bx + c = 0 are irrational.

  • Interesting; so this means that the graph of ax2+bx+cax^2 + bx + c never intersects the x-axis 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. b24ac2a\frac{b^2 - 4ac}{2a}. No, b±b24ac2a\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. The part that matters is inside the square root. We just need b24acb^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,ca, b, c are odd integers. They can be negative or positive, I notice.

  • Well, an odd integer squared is still odd. So b2b^2 is odd. And 4ac4ac 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 1×0=01 \times 0 = 0, odd times even is even).

    • Okay, so b24acb^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 4ac4ac to equal 16, but both aa and cc 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 4ac4ac 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 4ac4ac, 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 4ac4ac.

    • 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 written-up 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 499=4049 - 9 = 40 is not odd.

  • Perhaps we have to use the fact that b2b^2 is odd. But it's troubling that 4040 is not odd in the example above.

    • Why doesn't the example above work? That is, why can't we have b2=49b^2 = 49 and 4ac=404ac = 40? Well, aa would need to be 55 and cc would need to be 22 for that to work, but cc can't be 22. So we're saved, but how are we saved in general?
  • Okay, let's dive into the algebra. bb is odd, so say it's 2k+12k + 1.

    • Then b2=(2k+1)2=4k2+4k+1b^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... 4k2+4k4k^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,ca, c such that 4ac=4k2+4k4ac = 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 b24acd2b^2 - 4ac \ne d^2, where a,b,ca, b, c are odd integers and dd is any integer.

    • We can rewrite this into b2d2=4acb^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 4ac4ac" (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=1a = c = 1)? Well, no, as I laid out above there are no consecutive squares distance 4 apart, and the non-consecutive ones are too far apart to be 4 apart.

    • Can we just do odd and even? We know b2b^2 is odd, so subtract another square from it. If we subtract an even square, the result is odd and can't be 4ac4ac. If we subtract an odd square, things get more interesting...

    • Can we guarantee anything about relative size of bb and dd? I don't think so...

  • Subtract an odd square from an odd square. Say, 4k2+4k+1(4j2+4j+1)4k^2 + 4k + 1 - (4j^2 + 4j + 1). The 11s go away and we can take out a 4, so 4(k2+kj2j)4(k^2 + k - j^2 - j). We want to show this can't be 4ac4ac?

    • Well, k2+kk^2 + k is even regardless of what kk is, since it's odd plus odd. And same goes for j2+jj^2 + j. So it's 4 times an even minus an even = an even; a poor fit for 4ac4ac, I suppose.
  • This seems like it. But how to we justify that this cannot be 4ac4ac?

    • Oh, we can simply divide by 44. So, when you divide 4ac4ac by 4 the result is odd; when you divide 4(k2+kj2j)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 b24acb^2 - 4ac is not a perfect square. We proceed by contradiction. Suppose it were; that is, suppose b24ac=d2b^2 - 4ac = d^2, where dd is any integer. Rearrange: b2d2=4acb^2 - d^2 = 4ac. There are two cases to consider: if dd is even, and if dd is odd.

  • If dd is even, then d2d^2 is even. But then the left-hand side is odd (odd - even) and the right side is even, so there is a contradiction.

  • If dd is odd, then write 2k+12k + 1 for bb and 2j+12j + 1 for dd, where j,kj, k are any integer. Expanding, we get

b2d24k2+4k+1(4j2+4j+1)4(k2+kj2j)=4acb^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; k2+kj2j=ack^2 + k - j^2 - j = ac

  • Now, observe that the quantity k2+kk^2 + k must be even, since k2k^2 has the same parity as kk. Same with j2jj^2 - j. Therefore left-hand side k2+kj2jk^2 + k - j^2 - j is even.

  • But the right-hand side acac 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.