Let $f \in \mathbb{Q}[x]$.


Let $S(f)$ denote the number of distinct linear factors
of $f(x)-f(y)$.


$S(f)$ is bounded by $\deg(f)$.



> 
> Q1 Is $S(f)$ bounded by constant?
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> Q2 Is it possible $S(f)>2$ for arbitrary large degree of $f$?
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> 
> 


This is related to the number of ways of polynomial
represents an integer.



How many the distinct linear factors of $f(x)-f(y)$ can be for f in Q[x]?
$S(f) \leq 2$ for all nonconstant $f$.


By context, I understand that you mean factorization in $\mathbb{Q}[x,y]$. If you factor in $\mathbb{C}[x,y]$ then, of course, $x^n-y^n$ is a product of lines. (Daniel McLaury, in comments, mentions a third option, that $y$ might be a fixed integer. I don't know about this.)


Assume $f$ is not constant. Then clearly $x$ and $y$ do not divide $f(x)-f(y)$. Any other linear function can be written as $y-\lambda(x)$ for some affine linear transformation $\lambda(x) = mx+b$ with $m \neq 0$. If $y-\lambda(x)$ divides $f(y)-f(x)$ then $f(\lambda(x)) = f(x)$ for all $x$. Thus, $f(x) = f(\lambda(x)) = f(\lambda(\lambda(x)) = \cdots$. If $\lambda$ is not of either the form $\lambda(x) = x$ or $\lambda(x) = c-x$, then iteration of $\lambda$ has infinite orbits, so this implies $f$ is constant, a contradiction. (If we were factoring over $\mathbb{C}[x,y]$, we'd have to consider $m$ a root of unity other than $\pm 1$.)


Moreover, if $y-(c\_1-x)$ and $y-(c\_2-x)$ both divide $f(y)-f(x)$, then $f(x) = f(c\_1-x) = f(c\_1-c\_2+x) = \cdots = f(k(c\_1-c\_2)+x)$ for all $k$, so we can have $y-(c-x) | f(y)-f(x)$ for at most one $c$.


Thus, the most that can divide $f(y)-f(x)$ is $y-x$ and $x+y-c$ for one value of $c$, and $S(f) \leq 2$.


You might have intended to count factors with multiplicity, but this doesn't matter. Let the leading term of $f(x)$ be $f\_n x^n$. Then the degree $n$ part of $f(y) - f(x)$, namely $f\_n (x^n-y^n)$ is divisible by $x-y$ and $x+y$ only once, so $x-y$ and $x+y-c$ can each divide $f(x)-f(y)$ only once.


