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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 $x$.

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1 Answer 1

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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 $2^{y}$

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  • $\begingroup$ Is your approach gives all the solutions!. Or you just give some particular ones. $\endgroup$
    – Safwane
    Commented Jul 1, 2022 at 14:32
  • $\begingroup$ These are not all solutions. You can find all solutions by actually giving the question a moment's thought. $\endgroup$
    – Servaes
    Commented Jul 3, 2022 at 11:10

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