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Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $\frac{a^2 + b^2}{ ab + 1}$ is the square of an integer.

If we just divide $a^2+b^2$ with $ab+1$, we get a remainder in terms of $a$ and $b$. Since it is given that $ab+1$ divides $a^2+b^2$, we can equate the remainder to zero and get a relation between $a$ and $b$, which is: $a=b^3$ or $b=a^3$ , which on substitution gives us $b^2$ or $a^2$ hence proving what was to be proved.

Especially Lime
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Hubris
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  • Given that this was problem 6 of the 1988 IMO in Australia, a problem that was long considered the hardest one to ever appear in the olympiad, I don't think the solution is that simple. – Arthur Nov 23 '17 at 15:23
  • Yes, exactly, I'm trying to figure out flaws if any.. – Hubris Nov 23 '17 at 15:25
  • That being said, this problem is usually the example problem if you google the technique called "Vieta jumping". – Arthur Nov 23 '17 at 15:26
  • Yes, the vieta jumping, I've read about it, but I was unaware of it when I came across this question.. and all about this question on the internet has forced me to wonder if there's something fundamentally wrong with the approach I've attempted from. – Hubris Nov 23 '17 at 15:29
  • If you customize the quotient you would just harvest a random case of the general theorem, you would probably prefer to come through it by negation. – Abr001am Nov 23 '17 at 15:34
  • Duplicate of https://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer – Shri Nov 23 '17 at 15:35
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    @shrinit Just because they revolve around the same problem, doesn't mean they are duplicates. This question is about the flaws of one specific proof, and your linked question does not address this. – Arthur Nov 23 '17 at 15:39
  • Google after Vieta jumping – Dr. Sonnhard Graubner Nov 23 '17 at 15:43
  • Oh! my apologies.@Arthur – Shri Nov 23 '17 at 15:43
  • As I said We assume that there exist one or more solutions to the given condition for which k is not a perfect square What' is called vieta jumping is just a proof by contradiction. – Abr001am Nov 23 '17 at 15:51

2 Answers2

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Just because $a$ and $b$ are integers so that $ab+1$ divides $a^2+b^2$, that doesn't mean that the expression you get when you carry out the polynomial division $(a^2+b^2)\div(ab + 1)$ is nice in any way. After all, if it were a nice expression in $a$ and $b$ for some $a$ and $b$, then necessarily it would be a nice expression in $a$ and $b$ for all $a$ and $b$.

Also, you can't just equate the remainder with $0$. Take $\frac{a + 2}{a}$, for instance. Polynomial division gives a remainder of $2$, or possibly $\frac2a$, depending on what you mean by "remainder". Am I supposed to believe that $\frac{a+2}{a}$ is an integer as long as $2 = 0$, or $\frac2a = 0$? That doesn't bring me any closer to a solution.

Finally, I'm not sure that I know how to carry out the polynomial division $(a^2+b^2)\div(ab + 1)$. Are you certain that you do?

Arthur
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  • When a number divides another, in this case the numerator and the denominator are two integers, should mean that the remainder is zero isn't it? – Hubris Nov 23 '17 at 15:35
  • Yes, you should get remainder zero, but the remainder of polynomial division is not the same as the remainder you get when you do integer division after insertion. See my $\frac{a+2}a$ example again. – Arthur Nov 23 '17 at 15:37
  • multiply ab+1 with a/b or b/a, and add b^2 - (a/b) or a^2 - (b/a) to it respectively, you will end up with a^2 + b^2 , here, b^2 - (a/b) or a^2 - (b/a) is the remainder respectively... sorry for typing this way.. – Hubris Nov 23 '17 at 15:41
  • @Arthur I don't see what's wrong with your $\frac{a+2}{a}$ example. Equating the remainder to $0$, you get $\frac{a+2}{a}$ is an integer iff $2\equiv 0$ mod $a$, i.e. iff $a$ divides $2$, which is correct. The problem is just that polynomial division on the two expressions in OP doesn't go anywhere. – Especially Lime Nov 23 '17 at 15:44
  • I have equated the remainder to be zero since because it is "given" that a^2+b^2 is divisible by ab+1, in other words, "ab+1 divides a^2+b^2" – Hubris Nov 23 '17 at 15:48
  • Well, that's not equating the remainder to zero, that's making sure the remainder term from the polynomial division becomes an integer when you substitute. That is a correct approach, but you ought to be precise about what you do. That being said, how do you carry out the polynomial division. I would love to hear it, since I have no idea how it should be done. – Arthur Nov 23 '17 at 15:50
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    Why would you multiply $ab+1$ with $a/b$ or $b/a$? How do we know those quantities are integers? – G Tony Jacobs Nov 23 '17 at 16:06
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I don't know how you have divided $a^2+b^2$ by $ab+1$, or what remainder you've gotten. But your proof is certainly not correct, since you have "proved" that one number is the cube of the other, which is not necessarily true, e.g. take $a=8$, $b=30$.

Especially Lime
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  • polynomial division is what I've done.. but 8 and 30 hmm.. – Hubris Nov 23 '17 at 15:30
  • How do you do polynomial division? The quotient is about $a/b+b/a$ but since that isn't an integer this doesn't help. – Especially Lime Nov 23 '17 at 15:40
  • That's exactly where I've gone wrong.. my proof is valid only for the cases for which b perfectly divides a or a perfectly divides b, that should explain why 8 and 30 does not work. btw, I haven't divided all the way until you get a/b+ b/a as the remainder, but stopped when b^2- (a/b) or a^2-(b/a) is the remainder, (which I haven't further divided).. – Hubris Nov 23 '17 at 16:04
  • @Beast, if you edit your question to show us the steps of your polynomial division, we can probably help you spot the mistake. – G Tony Jacobs Nov 23 '17 at 16:05