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 , , and are odd integers, then the roots of are irrational.
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Interesting; so this means that the graph of never intersects the x-axis at an integer point.
- Not sure how to proceed from that; maybe we need to do something algebraic.
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Oh, obviously we can use the quadratic formula. . No, . The part that matters is inside the square root. We just need not to be a perfect square; if it isn't a perfect square, then neither will the roots be.
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Okay; we have that are odd integers. They can be negative or positive, I notice.
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Well, an odd integer squared is still odd. So is odd. And is even, since it's an even number times an odd number times an odd number.
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Is that right? Yes, it is, because evenness spreads like poision during multiplication (remember this via , odd times even is even).
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Okay, so is an odd minus an even, so it's odd.
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Wait; this doesn't get us anywhere. I don't care whether it's odd or not; I care whether it's a square.
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Unsure what to do.
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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.
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What do I have to subtract? Right; 16.
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Okay; that won't work. I have to get to equal 16, but both and are odd whereas I need them both to be 2 in order to get 16.
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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 from and the small square I'm trying to hit.
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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 , I've already seen that.
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Can I subtract to get to 9? No, I'd have to subtract 27, and that doesn't work either.
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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 .
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IS THAT IT??
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Right; when you list squares they look like this: 0, 1, 4, 9, 16 ... and the deltas look like this: 1, 3, 5, 7 ...
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It's all the odd numbers.
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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.
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I don't think so; for example is not odd.
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Perhaps we have to use the fact that is odd. But it's troubling that is not odd in the example above.
- Why doesn't the example above work? That is, why can't we have and ? Well, would need to be and would need to be for that to work, but can't be . So we're saved, but how are we saved in general?
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Okay, let's dive into the algebra. is odd, so say it's .
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Then .
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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?
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First one seems easier. But wait; how to subtract a square out?
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Okay, let's try the second one.
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Hmm, how do I subtract an even number out? I feel like that trailing 1 is giving me what I want. But... 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 such that , but I am at a loss as to why.
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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.
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We need to show that , where are odd integers and is any integer.
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We can rewrite this into 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 " (the product of two odd integers and 4).
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What can we say about the differences of squares? Are they ever just 4 (so )? 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.
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Can we just do odd and even? We know is odd, so subtract another square from it. If we subtract an even square, the result is odd and can't be . If we subtract an odd square, things get more interesting...
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Can we guarantee anything about relative size of and ? I don't think so...
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Subtract an odd square from an odd square. Say, . The s go away and we can take out a 4, so . We want to show this can't be ?
- Well, is even regardless of what is, since it's odd plus odd. And same goes for . So it's 4 times an even minus an even = an even; a poor fit for , I suppose.
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This seems like it. But how to we justify that this cannot be ?
- Oh, we can simply divide by . So, when you divide by 4 the result is odd; when you divide by 4 we know that the result is even. Hence they are different.
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(Took a break)
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Great, I think we can write up the final solution now.
Solution
By the quadratic formula, we just need to prove that is not a perfect square. We proceed by contradiction. Suppose it were; that is, suppose , where is any integer. Rearrange: . There are two cases to consider: if is even, and if is odd.
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If is even, then is even. But then the left-hand side is odd (odd - even) and the right side is even, so there is a contradiction.
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If is odd, then write for and for , where are any integer. Expanding, we get
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Divide both sides by 4;
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Now, observe that the quantity must be even, since has the same parity as . Same with . Therefore left-hand side is even.
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But the right-hand side 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.