Let $a>2$ and $b>2$ two strictely positive integers. Let us consider the following quantities: $$ f= \dfrac{xy+ay+a^{2}}{by}$$ $$g=a(y+a)\dfrac{xy+ay+a^{2}}{by^{2}x}$$ $$h=\dfrac{y+a}{b}$$

My question is:

Can we find integers $x$ and $y$ (not necessarly positive) such that $f,g,h$ are strictely positive integers. Or at lest how one can proves that they are exist.

  • $\begingroup$ Are you sure there are no other restrictions? It's too easy. Let $b = x = y =1$ and all your expressions are positive integers. $\endgroup$ – Tito Piezas III Jan 13 '18 at 17:54
  • $\begingroup$ @Tito: we could have $b > 1$ as $b$ is arbitrary $\endgroup$ – Ashwin Iyengar Jan 13 '18 at 17:55
  • $\begingroup$ @TitoPiezasIII: question edited. $\endgroup$ – China Jan 13 '18 at 17:59
  • $\begingroup$ @Ze2: If I use $b=2$ and $x\neq y$ it is still possible. $\endgroup$ – Tito Piezas III Jan 13 '18 at 18:08
  • $\begingroup$ @TitoPiezasIII: the nice answer is when $x$ and $y$ depend mainly on $a$ and $b$. $\endgroup$ – China Jan 13 '18 at 18:16

If we let,

$$a = (bk-1)y,\quad x = by$$

then the three polynomials lose their denominators,

$$(bk^2-k+1)y,\quad (bk-1)(bk^2-k+1)ky,\quad ky$$

and are positive integers if $b,k,y$ are positive integers.

  • $\begingroup$ This doesn't work: what if $a = 3$ and $b = 10$? Then how do you find $k$ and $y$ positive integers making $a = (bk-1)y$? $\endgroup$ – Ashwin Iyengar Jan 14 '18 at 16:06
  • $\begingroup$ @AshwinIyengar: The free parameters are $b,\,k,\,y$. Once chosen, they determine $a,x$. $\endgroup$ – Tito Piezas III Jan 14 '18 at 16:18
  • $\begingroup$ The first thing that is written in the question is "let $a > 2$ and $b > 2$ be strictly positive integers", which, I would think, implies that $b$ isn't free. $\endgroup$ – Ashwin Iyengar Jan 14 '18 at 18:06

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