Let us consider the following Diophantine problem:
Find all positive integer solutions verifying the two conditions:
(1) $(x+1)^2$ is a multiple of $2^{y}$
(2) $2^{y}≤x<2^{y+1}$
where $y$ is a fixed positive integer.
Context of the question: Let us consider the following differential equation:
$$z′=f(z,t)$$
where $f$ is a continuous function and $t,z∈ℝ$
When one search for the number of limit cycles of this equation, the above conditions holds. Now, the problem: Given $y$, can we decide whether the above conditions has a solution in positive integers $x$. In particular, I am interested on the cases where the number of limit cycles is zero, i.e., the above conditions has no solutions in positive integers $x$.
Consider $x=2^y+2n-1$, where $2n-1<2^{y+1}-2^{y}$, then $(x+1)^2$ is a multiple of $2^y$ only when $n=2^{y-1}$. If $x$ is even, then no solutions can exist because it leads to the conclusion that an odd number is a multiple of an even number, i.e., the odd number $(x+1)^2$ is a multiple of the even number $2^{y}$