How To prove Any Change to $v=a\cdot y+b$ maks $y=(a)^{−1}\cdot (v−b)$ Uni. random value This question is related to data integrity. 

Let a finite field be $\mathbb{F}_p$, where $p$ is a prime number.

I have a fixed value $y$ and two uniformly random values $a$ and $b$.

Hypothesis: $a,b,y,v \neq0$



I consider the server as an attacker who may modify the value $v$. The change to value $v$ may apply a clever change to value $y$ (see below).

I compute $v=a\cdot y+b$, and send it to the server. The server can do any change to $v$. So it computes a function of $v$, as $f(v)$. 

Assume that I can magically detect if $y$ becomes a uniformly random value. So as long as the server cannot change $y$ to an arbitrary value $y'$ the security property is satisfied.

I can download $f(v)$ and remove the random values to obtain $y$, so I can do: $y=(a)^{−1}\cdot (f(v)−b)$



Question: How can I prove that ANY change to $v$ makes $y$ a uniformly random value?



To clarify I want to make sure the server cannot modify $v$ to apply an arbitrary change to $y$. In other word, my security needs would be satisfied if the server only can change $y$ to a uniformly random $y'$ (by changing $v$)
Well, there is not much to prove.


  The server can do any change to $v$.


So actually, this means the server with throw away your $v$ and just choose a new random number. Or alternatively, he chooses $w$ uniformaly random and sets $v'=v+w$, which is now also uniformly random.

When you get back that value, you have your (for you known/fixed) values $a,b$ and do your calculation $y' = a^{-1}(v'-b)$. With he existence of $a^{-1}$ (i.e. $a\neq0$), this is also uniformly distributed for fixed values $a,b$.