1

Let $a,b\in+\mathbb Z$ such that $\dfrac{a^2+b^2}{ab+1}=k.$

Prove that $k$ is the square of an integer when $$a^2+b^2$$ is divisible by $ab+1$.

Soln. For $k$ to be a perfect square, $\dfrac{a^2 + b^2}{1+ab}$ must be in the form $\dfrac{mp^{n+2}}{mp^n}$ where $p\in +\mathbb Z$ and $n,m\in\mathbb Q.$

Hence $\dfrac{a^2+b^2}{ab+1} = \dfrac{mp^{n+2}}{mp^n}.$ On comparison of numerator and denominator we have $$ a^2 + b^2 = mp^{n+1} \tag 1 $$ and $$ ab+1 = mp^n. \tag 2 $$ Solving for $a$ and $b$ we get \begin{align} b & =\sqrt{\frac{mp^{n+2} \pm \sqrt{m^2 p^{2n+4} -4m^2 p^{2n} +8mp^n -4}}{2}} \\[6pt] \text{and } a & = (mp^n -1)\sqrt{\frac{2}{mp^{n+2} \pm \sqrt{m^2 p^{2n+4} -4m^2 p^{2n} +8mp^n -4}}} \end{align} so $k$ will be perfect square when $a$ and $b$ can be expressed in this form where $p\in +\mathbb{Z}$ and $m,n\in \mathbb{Q}.$

Can this be a proof?

Daniel
  • 77
  • 1
    The problem asks to show that $k$ is always an integer's square. You just showed it's a square if some very complicated condition is true. – aschepler Jul 22 '20 at 16:52
  • 2
    Writing everything in italics vitiates the whole point of using italics for math expressions. – saulspatz Jul 22 '20 at 16:52
  • There seems to be some other implied or missing condition? Like maybe $k$ is an integer? – aschepler Jul 22 '20 at 16:55
  • 1
    I wonder if I've ever seen a worse way of using MathJax. I cleaned up a little bit of it. – Michael Hardy Jul 22 '20 at 17:01
  • @Michael Hardy Sorry for poor latex command actually I use the website match.io to convert mathematical symbol in Latex form. – Daniel Jul 23 '20 at 03:30
  • @aschepler maybe this can be called conditional proof en.m.wikipedia.org/wiki/Conditional_proof . This question has been taken from IMO ques 6 1988. Moreover n and m could be any rational no. which will satisfy that a , b are integers . – Daniel Jul 23 '20 at 06:11
  • 1
    Except it's not true as stated. If we take $a=1, b=2$, then $k=\frac{5}{3}$, which is definitely not the square of an integer. The original wording is "such that $ab+1$ divides $a^2+b^2$", which does mean $k$ must be an integer. – aschepler Jul 23 '20 at 10:50

0 Answers0