Question

How to estimate the probability of co-occurrence of the positive integers $c_i$ and $d_i$, $1 \leq i \leq t$ drawn from the uniform range $1$ to $2^k-1$, such that $\Sigma^t_{i=1} c^2_i = \Sigma^t_{i=1} c_i\times d_i$ and $\forall i, c_i\neq d_i$ hold?

$\forall i, c_i$ and $d_i$'s are drawn from the same distribution, but each pair $(c_i, d_i, \forall i \neq j)$ is drawn from different distribution.

To simplify the problem for the case $t=2$, it is the probability of co-occurrence of four positive integers $c_1, c_2, d_1$ and $d_2$ such that $c^2_1 + c^2_2 = c_1\times d_1 + c_2\times d_2 $ and $c_1\neq d_1$ and $c_2\neq d_2$ hold. The pairs $(c_1,d_1)$ and $(c_2,d_2)$ are chosen from two different distributions of the same range $1$ to $2^k-1$.

Answer 1

An approach for the case $t=2$. There are $N^4$ ways of choosing $a,b,c,d$, all between 0 and $N-1$. You want to know how many of them result in $a^2+b^2=ac+bd$. Let's take the case where $\gcd(a,b)=1$ and $a\lt b$. Then you must have $c=a+br$ and $d=b-ar$ for some $r$. This $r$ can't exceed $b/a$, nor can it exceed $(N-a)/b$. So, given $a,b$, the number of pairs $c,d$ is just the smaller of $b/a$ and $(N-a)/b$, suitably rounded. Now you have to sum that over all pairs $a,b$ with $\gcd(a,b)=1$ and $a\lt b$. Then you have to work out what modifications you need to handle the case $\gcd(a,b)=e\gt1$.

EDIT: Fixing $t$ but letting $N$ go to infinity, the probability must go to zero, since, when you've picked all the variables but one, there's only one way to choose that last variable (unless its partner is zero), so the probability should never exceed $2/N$. But maybe I misunderstand the question.