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Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$\frac{f(x)}{y} - \frac{f(y)}{x} > 2$$|\frac{f(x)}{y} - \frac{f(y)}{x}| > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

for all $x, y$ in the domain of $f$.

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$\frac{f(x)}{y} - \frac{f(y)}{x} > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

for all $x, y$ in the domain of $f$.

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$|\frac{f(x)}{y} - \frac{f(y)}{x}| > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

for all $x, y$ in the domain of $f$.

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

edited in response to a clarification
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Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$\frac{f(x)}{y} - \frac{f(y)}{x} > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

for all $x, y$ in the domain of $f$.

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$\frac{f(x)}{y} - \frac{f(y)}{x} > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$\frac{f(x)}{y} - \frac{f(y)}{x} > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

for all $x, y$ in the domain of $f$.

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

added another constraint (thanks to Gerhard Paseman)
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