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I have tried using Induction method but I am unable to resolve it to a single variable.

Also, $x$ & $y$ are positive integers.

$f\:=\:\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}$

Edit : This one is simpler than the linked question and requires less steps.

tommy vichi
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    Possible duplicate of Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$. – FredH Apr 04 '19 at 23:40
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    Hint: Start by proving that $3\mid x^2+y^2+1$ if and only if $3\not\mid xy$. – Barry Cipra Apr 04 '19 at 23:40
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    @FredH: This one is much simpler because we do not have to show $f=3$, just that it is a multiple of $3$. That one proves this one, but we do not need so much technique here. – Ross Millikan Apr 04 '19 at 23:41
  • @BarryCipra I just answered this at another question today. The general result is not related to factoring, it is an inequality. https://math.stackexchange.com/questions/3174167/understanding-a-proof-that-if-xy-divides-x2y21-for-positive-integers-x/3175063#3175063 – Will Jagy Apr 04 '19 at 23:46
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    @BarryCipra Sorry but the hint is not helping. – tommy vichi Apr 04 '19 at 23:48
  • @WillJagy, interesting, the question you linked to is a duplicate of the one in FredH's link. – Barry Cipra Apr 05 '19 at 00:13
  • @BarryCipra you have a point. There is an answer here that may be correct, and is certainly short and simple. The full question has been posted many, many times here. It is the "constant descent" example at https://en.wikipedia.org/wiki/Vieta_jumping – Will Jagy Apr 05 '19 at 00:14
  • @WillJagy Can you please elaborate the given solution. – tommy vichi Apr 05 '19 at 00:16
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    @tommyvichi, it would help if you provided some additional context regarding what you've tried, along with how you encountered the problem. It looks to me like a problem in a course on elementary number theory rather than something in algebra/precalculus. – Barry Cipra Apr 05 '19 at 00:22
  • Please try using AM-GM bounding, it just struck me now. Can you prove it using that since we are only using integers. – tommy vichi Apr 05 '19 at 00:25
  • @Will Your link is a different question. This is much simpler, e.g. see the one-line proof in my answer. It would be helpful to remove (and/or repost) your comment to avoid any further confusion. – Bill Dubuque Apr 05 '19 at 01:11
  • I will provide the full context and reask it – tommy vichi Apr 05 '19 at 19:10

3 Answers3

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$f, x, y$ are integers. (The proposition is not true if we allow $x,y$ to be rationals. e.g. let $x = y = \frac 13$ then $f = 11$)

$f = \frac {y}{x} + \frac {x}{y} + \frac {1}{xy} \implies xyf = x^2 + y^2 + 1$

$xyf \equiv x^2 + y^2 + 1 \pmod{3}$

If $x$ or $y$ are divisible by 3 then $xyf \equiv 0 \pmod 3.$

But then, it would not be possible for $x^2 + y^2 + 1 \equiv 0 \pmod 3$

If neither $x$ nor $y$ is $0$ modulo $3$ then $x^2 + y^2 + 1 \equiv 0 \pmod 3$

In which case $f\equiv 0 \pmod 3$

A quick primer on the field of integers mod $p$

We can think of modulo $p$ as the remainder we get when we divide by $p.$

If a number is equivalent to 0 mod p then it is divisible by $p.$

Modular arithmetic preserves addition and multiplication. That is, if we add to numbers and evaluate modulo p, we can just as easily evaluate the numbers modulo p before we add them. And, if we multiply integers and evaluate mod p, we can first evaluate modulo p and then multiply. If we get a result bigger than p we can evaluate that result modulo p.

We can use this to dramatically simplify problems, as there are only p cases to check for each variable in the equations, or for this problem 3 cases.

When we look at $x^2 + y^2 + 1$ we only have to consider: $1^2 \equiv 1 \pmod 3, 2^2 \equiv 1 \pmod 3$ and $0^2 \equiv 0 \pmod 3.$ If $x,y$ are not divisible by 3 then $x^2 + y^2 + 1 \equiv 0 \pmod 3$ If x or y are divisible by 3, then $x^2 + y^2 + 1 \equiv 1,2 \pmod 3$

If p is prime, the integers mod p form a ring. That is $ab\equiv 0\pmod p \implies$ p divides a or p divides b. Toward the end of the proof, we are using this property. We have shown that $xyf = x^2 + y^2 + 1 \equiv 0 \pmod 3$ and $x,y$ are not divisible by $3$ therefore $f$ is.

user317176
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Hint $\bmod 3\!:\,\ {-}1 \equiv x^2-fxy+y^2\equiv (x+fy)^2\ $ if $\, f\equiv \pm1,\,$ so $\,f\equiv 0$

Bill Dubuque
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Write it as $$ xyf = x^2 + y^2 + 1 $$ Compute both sides mod $3$, that is, for $x,y=0,1,2$. It turns out that there are two cases:

  • The LHS is zero and the RHS is not zero. This cannot happen.

  • The LHS is not zero and the RHS is zero. This implies that $f \equiv 0 \bmod 3$.

$$ \begin{array}{cccc} x & y & xy & x^2 + y^2 + 1 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 2 & 0 & 2 \\ 1 & 0 & 0 & 2 \\ 1 & 1 & 1 & 0 \\ 1 & 2 & 2 & 0 \\ 2 & 0 & 0 & 2 \\ 2 & 1 & 2 & 0 \\ 2 & 2 & 1 & 0 \\ \end{array} $$

lhf
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  • Thank you for your response. Can you please elaborate a little bit on how to compute mod 3 in case of these algebraic equations. – tommy vichi Apr 05 '19 at 01:04