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Let us consider the function:

$$f(x,y)=\dfrac{axy+by+cx+d}{rxy+sy+hx+w}$$

Here $f(x,y)$ is the division of two quadratic polynomials with the same degree

where $a,b,c,d,r,,s,h,w$ are non zero integers.

My question is:

(1) Given $a,b,c,d,r,,s,h,w$ as integers, can we find $x$ and $y$ rational numbers such that $f(x,y)$ is a strictely positive integer.

(2) Given $a,b,c,d,r,,s,h,w$ as integers, can we find $x$ and $y$ positive integers such that $f(x,y)$ is a strictely positive integer.

add

We can assume that $f$ is bijective with respect to $x$ if $y$ is a constant and bijective with respect to $y$ if $x$ is a constant.

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  • $\begingroup$ Are there any restrictions whatsoever on $(a,b,c,d,r,s,h,w)$? (For instance, we will obviously have a hard time if $a=b=c=r=s=h=0$, because then our choice for $(x, y)$ doesn't even matter.) $\endgroup$ – Lynn Jan 7 '18 at 14:27
  • $\begingroup$ @Lynn: No. Only they are non zero integers. $\endgroup$ – China Jan 7 '18 at 14:32
  • $\begingroup$ With bijectivity condition, it is easy. Fix some $x$, then choose $y$ such that $f(x, y) =1$ (you can because of surjectivity). Maybe the condition you want is "non constant" instead of "bijective"? $\endgroup$ – Andrea Marino Jan 12 '18 at 11:50
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The denominator can be the opposite of the numerator, so that the function is a negative constant. Take for instance $$\frac{xy+x+y+1}{-xy-x-y-1}$$ in this way $f(x, y) = - 1$ and you have no chance of finding such $(x, y) $!

Let me enounce a simple fact: For every fixed integers $u, v$, it exists an integer N of the same sign of $u$ such that for every $n$ of the same sign of $u$ with $|n|>|N|$ we have

$$un+v >0 $$

As a corollary, we have : If u, u' have the same sign, for every v, v' it exists n such that un+v, u'n+v' are positive. Proof. Take N, N' given by the previous fact: we know they have the same sign. Thus if we take a n of the same sign of N, N', but greater in modulus, we have $un+v, u'n+v'$ both positive.

Now i will show you that if $a, r$ have the same sign, the answer is affirmative in both cases. In fact, choose $x$ of the same sign of $a$, big enough such that both $ax+b, rx+s$ are positive. Then you can find $y$ such that both the linear forms $(ax+b) y + (cx+d), (rx+s) y+(hx+w) $ are positive, because the coefficient of $y$ have the same sign. In conclusion, the quotient

$$\frac{axy+by+cx+d}{rxy+sy+hx+w} $$

Is positive!

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  • $\begingroup$ @ Andrea Marino: Can you explain on more detaills. $\endgroup$ – China Jan 7 '18 at 15:24
  • $\begingroup$ I have filled in more details! Is it clearer now? $\endgroup$ – Andrea Marino Jan 8 '18 at 17:49
  • $\begingroup$ @ Andrea Marino: But the last expression is not necessarly a positive integer. $\endgroup$ – China Jan 9 '18 at 12:42
  • $\begingroup$ hahah, ok, I didn't read that condition. The problem became more interesting! Let me make some changes. $\endgroup$ – Andrea Marino Jan 9 '18 at 22:45
  • $\begingroup$ I have a solution to both the questions I will write in a few days! I have a complete characterization of point (1), while just a counterexample to (2) more refined than the constant function. $\endgroup$ – Andrea Marino Jan 15 '18 at 0:50

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