How many integers are there?

$4n-1$ must be a perfect square to satisfy the condition of the problem. So, let $4n-1=a^2$ , where $a$ is an integer.

Then $n=\dfrac {a^2+1}{4}$

Hence $a$ must be an odd number. Let $a=2b+1$

Then $n=\dfrac {4b(b+1)+2}{4}$

Now this number can never be an integer, whatever be the value of $b$

Hence the answer is $\boxed 0$ .

$\begin{aligned} \text{Any perfect square} &\equiv 0 \text{ or } 1 \pmod {4} & \small{\color{#D61F06}{\text{(See wiki)}}} \\ & \to \text{No solution} \\ \end{aligned}$ $\small{\color{#3D99F6}{\text{Link to Wiki}}}$