Let  $f\left(x\right)=\frac{\sin \pi x}{x^2},x>0$
Let  $x_1<x_2<x_3<\mathrm{.......}{x}_n<\mathrm{...}$ be all the points of local maximum of f. 
and  $y_1<y_2<y_3<\mathrm{.......}{y}_n<\mathrm{...}$ be all the points of local minimum of f.

$f\left(x\right)=\frac{\sin \pi x}{x^2}\Rightarrow f^{\text{'}}\left(x\right)=\frac{\pi x^2\cos \pi x-2x\sin \pi x}{x^4}=\frac{2\cos \pi x(\frac{\pi x}{2}-\tan \pi x)}{x^3}$

$$\begin{gathered} f^{\text{'}}\left(x\right)=0\Rightarrow \cos \pi x=0 or \frac{\pi x}{2}=\tan \pi x \\ \Rightarrow \pi x=(2n+1)\frac{\pi }{2} or \frac{\pi x}{2}=\tan \pi x \\ x=\frac{(2n+1)}{2},n\in 1 \end{gathered}$$
From graph, we can see that  $\forall x=\frac{2n+1}{2}$
 
$\Rightarrow f^{\text{'}}\left(x\right)$  doesn’t change sign so these points are neither local maxima nor local minimum
 $$\begin{gathered} Similarly, \forall x:\frac{\pi x}{2}=\tan \pi x \\ Where y_n\in (2n-1,2n-\frac{1}{2})\forall n=1,2,\mathrm{3.....} \\ and x_n\in (2n,2n+\frac{1}{2})\forall n=1,2,\mathrm{3....} \end{gathered}$$

$x_{n+1}-y_{n+1}>1 and y_{n+1}-x_n>1\Rightarrow x_{n+1}-x_n>2$