1

I asked this question yesterday, and I think I have come up with an (albeit incomplete) proof of my own that I want to complete. The proof proceeds thus:

Given $x + y = xy$, we infer that $x, y$ are even. Hence, we can take $x = 2x_1$ and $y = 2y_1$ where $x_1, y_1 > 1$ [1]. Substituting this, we get $2x_1 + 2y_1 = 4x_1y_1 \implies x_1 + y_1 = 2x_1y_1$. Similarly, we see that $x_1, y_1$ are even [2] as well, which means $x_1 = 2x_2$ and $y_1 = 2y_2$. Substituting again, we get $x_2 + y_2 = 4x_2y_2$. This repeats infinitely, implying that $x, y$ are divisible by all powers of two. We know this is true only for $(0, 0)$. Hence, our only solutions are $(2, 2)$ and $(0,0)$.

I have the following problems with this proof, the list index refers to parts of the proof:

  1. Am I allowed to make such an assumption?
  2. What can I do about the case where $x_n, y_n$ are both odd? I've tried taking $x_1 = 2x_2 + 1$ and $y_1 = 2y_2 + 1$, but I get a contradiction ($x_2 + y_2 = -4x_2y_2$). So do I conclude they can't both be odd?

I am aware this is a triviality to solve using algebra, but I really want to understand the concept of infinite descent through this question.

  • 3
    Why not just using $$(x-1)(y-1)=xy-x-y+1=1$$ because of the given equation ? The factors $x-1$ and $y-1$ must both be $-1$ or $+1$ . That's it. – Peter Sep 02 '22 at 16:22
  • 1
    The second time that you conclude that both are even doesn't follow, since that equation has the solution $(1,1)$. – plop Sep 02 '22 at 16:23
  • 1
    @Peter please check the last part of the question. –  Sep 02 '22 at 16:25
  • @user85667 i am aware, i just can't figure out how i can include it in this proof –  Sep 02 '22 at 16:25
  • 1
    It is not a proof, if it contains an implication that is invalid. – plop Sep 02 '22 at 16:27
  • 1
    yes, that is precisely why i labelled it incomplete @user85667 should i have been clearer? –  Sep 02 '22 at 16:30
  • @user85667 $;(1,1);$ isn't a solution of this equation... – DonAntonio Sep 02 '22 at 16:37
  • @DonAntonio You don't know what you are talking about. Read first. – plop Sep 02 '22 at 16:38
  • For easy solutions see also here. – Dietrich Burde Sep 02 '22 at 16:45
  • 1
    Expanding on the comment of user85667, every time after the first that you assert that $x_i,y_i$ must be even is wrong. From the equations per se, you can only conclude that $x_i,y_i$ have the same parity. You should realize something is wrong when your argument doesn't allow only $0,0$, but also $2,2$. Those aren't divisible by every power of $2$ as you asserted was necessary. – Keith Backman Sep 02 '22 at 17:06
  • Assumption 1. Either $x_1 < y_1$ or $x_1 = y_1$ or $x_1 > y_1$. If $x_1 > y_1$ then, as both addition and multiplication are commutative, we can just relabel to variables and rename $x_1$ as $y_1$ and $y_1$ as $x_1$ and we can say: we can assume without loss of generality that $x_1 < y_1$ or $x_1 = y_1$. We often abbreviate "without loss of generality" as wolog. But you need to make a case if $x_1 = y_1$ and $4x_1 = 4x_1^2$ (in which case $x_1=1$ and $x=y=2$ is a solution). – fleablood Sep 02 '22 at 19:33
  • As for assumption 2: why would $a + b = 2ab$ imply $a,b$ are both even? $2ab$ is even and that means $a+b$ are the same parity but they could both be odd. – fleablood Sep 02 '22 at 19:35
  • Why not notice that if $x+y = xy$ and we assume $x\ne 0; y\ne 0$ ($x=0;y=0$ is a solution but that is the only solution where one is zero) then $1 + \frac yx = y\in \mathbb y$. That means $x|y$. But $\frac xy + x = x\in \mathbb Z$ so $y|x$. So $x = \pm y$. So we have either $x + x = x^2$ or $x-x = x^2$. So either $x=-y=0$ or $2x =x^2$ and $x = y = 2$. – fleablood Sep 02 '22 at 19:41
  • I just now posted an answer on your prior question which explains how to view this as a special case of (infinite) descent on integer points on conics - a method that works generally. This is rather trivial in this case, but far less trivial in other cases, e.g. see the linked post there on vieta jumping. I highly recommend that you learn these general methods to best understand these matters. – Bill Dubuque Sep 02 '22 at 19:50
  • "I've tried taking x1=2x2+1 and y1=2y2+1, but I get a contradiction (x2+y2=−4x2y2). " How so? If $x_1=2x_2 + 1$ and $y_1 = 2y_2 + 1$ we get $x_1 + y_1 = 2x_2 + 2y_2 + 2= 2(2x_2+1)(2y_2+1)=8x_2y_2 + 4x_2 + 4x_1 + 2\implies x_2+y_2 = 4x_2y_2 + 2x_2 + 2x_1$ and how is that a contradiction? Nor is $x_2 + y_2 =-4x_2y_2$ a contradiction [Note: $x_2 =y_2 = 0$ is the actual solution] although I do not see how you get that. This really isn't a good case for infinite regress unless you have some reason to say "whoa, these numbers are getting out of control" – fleablood Sep 02 '22 at 19:50
  • Btw the method in the first comment above can be viewed as a generalization of completing a square - see completing a product (or rectangle). – Bill Dubuque Sep 02 '22 at 20:16
  • @fleablood ignore that part, i was under the impression that for whatever reason i was limited to the natural numbers yesterday when i made the post. and for your previous comment, that is exactly how i solved it the first time too. as for the first assumption, i get what you're telling. for the second, note that i don't want to make such a claim, i just didn't know how i could deal with odd cases. now i realise that this is the same process as the cases where x and y are even, just with a negative sign. is this a correct realisation? –  Sep 02 '22 at 23:43

1 Answers1

2

The question admits of no solution via descent, as there is a single solution for $x,y>0$, not a series of hypothetical solutions that descend to one actual solution.

The starting equation is symmetrical in $x,y$. The trivial solution $(0,0)$ is available by inspection. For $x,y>0$, simple division by either $y$ or $x$ gives $$x=1+\frac{x}{y}\\ y=1+\frac{y}{x}$$ Since $x,y \in \mathbb Z$, it follows $\frac{x}{y},\frac{y}{x}\in \mathbb Z$

Together these imply both $x\ge y$ and $y\ge x$, whence $x=y$

Ergo $\frac{x}{y}=1,\frac{y}{x}=1$, and $x=2,y=2$

Keith Backman
  • 7,681
  • Incorrect - see this answer to the OP's prior question (which is where this post belongs, since this followup question is about the correctness of the OP's specific approach here). – Bill Dubuque Sep 02 '22 at 20:23