solve it #1

start of with the substitution $x=u-\dfrac{3}{2}$ we get that $(u-\dfrac{3}{2})(u-\dfrac{1}{2})(u+\dfrac{1}{2})(u+\dfrac{3}{2})=y^2$ upon expansin $u^4 -\dfrac{5}{2}u^2+\dfrac{9}{16}=y^2$ multiplying both sides with 16 $u^4-40u^2 +(9-16y^2)=0$ using the quadratic formula $u^2=\dfrac{5\pm 4\sqrt{y^2+1}}{4}$ we see that for u^2 to be rational, y must equal 0 $u^2 =\begin{array}{c}=\dfrac{5+4}{4}\quad or\quad \dfrac{5-4}{4}\\ =\dfrac{9}{4}\quad or\quad\dfrac{1}{4} \end{array}$ which means u could be $u=\pm\dfrac{3}{2},\pm\dfrac{1}{2}$ now re-substitute and put these 4 values $x= \dfrac{3}{2}-\dfrac{3}{2},-\dfrac{3}{2}-\dfrac{3}{2},\dfrac{1}{2}-\dfrac{3}{2} -\dfrac{1}{2}-\dfrac{3}{2}$ $x=0,-3,-1,-2$ none of which are natural numbers. so the no.of solution is, hence $\boxed{\large{\color{#D61F06}{zer0}}}$