This question may seem to be related to Probability and Data Integrity but mine is much simpler and consideres a DIFFERENT problem.
Let a finite field be $\mathbb{Z}_p$, where $p$ is a prime number.
I have a fixed value $y_i$ and two uniformly random values $a_i$ and $b_i$.
I compute $v_i=a_i\cdot y_i + b_i$, and send it to the server. The server can do any change to $v_i$. Assume that I can magically detect if $y_i$ becomes a uniformly random value (see below how it might be possible)
I can download $v_i$ and remove the random valus to obtain $y_i$
Question: How can I prove that any change to $v_i$ makes $y_i$ a uniformly random value?
Imagine we have a polynopmial $P(x)=(x-\beta)\cdot g(x)$ define over the field. Then evaluate $P(x)$ at $n$ number of $x_i$'s. This gives us $y_i$'s. So if any subset of $y_i$ is replaced with a set of uniformly random values, with a high probability we cannot retrieve $\beta$ as a root of interpolating polynomial.
TBN: $a_i,b_i,y_i\neq0$
Let us first drop the subscripts, and let $z= a^{-1} \left( f ( a y + b) - b \right)$, where $f : \mathbb{Z}_p \to \mathbb{Z}_p$ is any change the computer makes to $ay+b$.
Then we want to show that given any $f$, $z$ is uniform distributed. Unfortunately, this is not true in general.
If you want a counterexample in $\mathbb{Z}_p$ for any $p$, we take $f$ to be the indicator of $\{0\}$, and then we have, we assume $a$ and $b$ independent, \begin{eqnarray} \mathbb{P}(z=0) &=& \mathbb{P}( a^{-1} ( f(ay+b) - b) = 0 ) \\ &=& \mathbb{P}(f(ay+b) -b = 0) \\ &=& \mathbb{P}(f(ay+b) = b) \\ &=& \mathbb{P}(f(ay+b) = 1 \textrm{ and } b=1) \\ &=& \mathbb{P}(a = -y^{-1} \textrm{ and } b=1 )\\ &=& \mathbb{P}(a = - y^{-1}) \mathbb{P}(b=1) \\ &=& \frac{1}{(p-1)^2} \end{eqnarray} So for $z$ to be uniform, we must at least have $(p-1)^2$ elements in $\mathbb{Z}_p$, which clearly is not possible for $p>2$.
If we have that $z \neq 0$, we have the constraint $ f(ay+b) - b \neq 0$.
We will show that this yields quite a constraint on the allowed functions $f$, and that for only one function this yields $z \in U(1,p-1)$ when $p>2$.
Let $y \in \mathbb{Z}_p$ be fixed and non-zero. Since $a$ is uniform and non-zero, we have that $ay \in \mathbb{Z}_p \backslash \{0\}$, where every element has a chance of $\frac{1}{p-1}$.
So given $v=ay+b$, we can find for all $\tilde{b} \in \mathbb{Z}_p \backslash \{ 0, v \}$ a $\tilde{a} \in \mathbb{Z}_p \backslash \{0\}$ such that $\tilde{a}+\tilde{b} = v$.
So if the computer knows $v$, it can only know that $b \neq v$, since $b \neq 0$.
So we must have $f(x) \in \{0,x\}$ for all $x \in \mathbb{Z}_p$.
If we have $v_i \in \mathbb{Z}_p \backslash \{0\}$ such that $f(v_i) = v_i$ for $i=1,\dots,k$ we find \begin{eqnarray} \mathbb{P}(ay+b = v_i) &=& \sum_{\tilde{b} \not\in \{0,v_i\} }\mathbb{P}(b = \tilde{b}) \mathbb{P}(a = y^{-1} ( v_i - \tilde{b})) \\ &=& \sum_{\tilde{b} \not\in \{0,v_i\}} \frac{1}{p-1} \frac{1}{p-1} \\ &=& \frac{1}{p-1} -\frac{1}{(p-1)^2} \end{eqnarray} For $ay+b = v_i$ we have $z=a^{-1}( f(ay+b) -b) = a^{-1}( ay+b - b)=y$. So we find \begin{eqnarray} \mathbb{P}(z=y) &\ge& \sum_{i=1}^k \mathbb{P}(ay+b = v_i) \\ &=& \sum_{i=1}^k \frac{1}{p-1} - \frac{1}{(p-1)^2}\\ &=& \frac{k}{p-1} -\frac{k}{(p-1)^2} \end{eqnarray} And for $k\ge 2$ we have that this is strictly greater than $\frac{1}{p-1}$, so $z$ is not uniform.
So assume there exists $v \in \mathbb{Z}_p \backslash \{0\}$ such that $f(v)=v$ and $f(x)=0$ for all $x \neq v$.
Then we have $\mathbb{P}(ay+b=v) = \frac{1}{p-1}-\frac{1}{(p-1)^2}$.
Furthermore, we have \begin{eqnarray} \mathbb{P}(- a^{-1}b=y) &=& \mathbb{P}(b = -ay ) \\ &=& \sum_{\tilde{a} \neq 0} \mathbb{P}(b = -\tilde{a}y ) \mathbb{P}(a = \tilde{a} ) \\ &=& \sum_{\tilde{a} \neq 0} \frac{1}{p-1} \frac{1}{p-1} \\ &=& \frac{1}{p-1} \end{eqnarray} We note that $-a^{-1}b=y$ implies $ay +b = ay -ay =0 \neq v$.
So we have \begin{eqnarray} \mathbb{P}(z=y) &=& \mathbb{P}(ay+b =v) + \mathbb{P}(-a^{-1} b =y)\\ &=& \frac{1}{p-1} - \frac{1}{(p-1)^2} +\frac{1}{p-1}\\ &=& \frac{2}{p-1} - \frac{1}{(p-1)^2}\\ &>& \frac{1}{p-1} \end{eqnarray} So we again have that $z$ is not uniform.
Thus the only remaining possible function is $f = 0$. This gives for $v \in \mathbb{Z}_p \backslash \{0\}$ \begin{eqnarray} \mathbb{P}(z =v ) &=& \mathbb{P}( -a^{-1}b = v) \\ &=& \sum_{\tilde{a} \neq 0 } \mathbb{P}(a = \tilde{a} ) \mathbb{P}(b = -v\tilde{a})\\ &=& \frac{1}{p-1} \end{eqnarray} So we have $z$ is uniform.