number of solutions to $(x+t)^a=x^a$ Let $\mathbb{F}_q$ with $q=p^m$ ($p$ is prime) denote a finite field. For any $t\in\mathbb{F}_q^*$ and a ratioanl number $a$, consider the equation
$$
(x+t)^a=x^a
$$
Let $A(t,a)$ denote the number of solutions to the above equation in $\mathbb{F}_q$. Then can we bound $A(t,a)$ by a constant depending only on $a$ (i.e. independent of $q$ and $t$)? Will the answer be different if we only consider the special case $m=1$ (i.e. $q=p$)? 
 A: Like KReiser I am worried about the meaning of $x^a$ for non-integer exponent $a$ in a finite field. On the other hand, if $a=r/s$ with integers $r,s$ then we can define $x^{r/s}$ as the unique solution $z\in\Bbb{F}_q$ of the equation
$$
x^r=z^s\qquad(*)
$$
when and only when $\gcd(s,q-1)=1$. Without the gcd-condition the equation $(*)$ may have no solutions at all, or it will have $\gcd(s,q-1)$ distinct solutions. This follows from cyclicity of the group $\Bbb{F}_q^*$.
So we need to assume that either $a$ is an integer or $a=r/s$ with $\gcd(s,q-1)=1$. We can make the following observations:


*

*When $\gcd(s,q-1)=1$ we have $(x+t)^{r/s}=x^{r/s}$ if and only if $(x+t)^r=x^r$, so we can ignore the denominator and $A(t,r/s)=A(t,r)$.

*For the same reason we have $A(t,ks)=A(t,k)$ for all integers $k$ and all integers $s$ such that $\gcd(s,q-1)=1$. 

*As a corollary of the previous observation we have $A(t,a)=A(t,\gcd(a,q-1))$ whenever $a$ is an integer. In other words, without loss of generality we can assume that $a\mid q-1$

*The equation $(x+t)^r=x^r$ can be viewed as a polynomial equation of degree $r-1$, so we have the trivial upper bound $A(t,r)\le r-1$. 

*The substitution $x=ty$ (recall that $t\neq0$) transforms $(x+t)^a=x^a$ to $t^a(y+1)^a=t^ay^a$ which is obviously equivalent to $(y+1)^a=y^a$. Consequently $$A(t,a)=A(1,a)$$ for all exponents $a$ and all $t\neq0$.


Edit: I misremembered, and thought that the equation occurs in the study of so called almost perfect non-linear mappings. But, there the equation of interest is $(x+t)^a=x^a+b$. Instead, we can  proceed from this point as pointed out by Jake Levinson (see comments below).
