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Why $y^2 = x^3 + 7$ has no integral solutions?

Consider the given equation...
$$y^2-7=x^3=(x^{\frac{3}{2}})^2$$
$$y^2-(x^{\frac{3}{2}})^2=7$$
If $x^{\frac{3}{2}}$ is a decimal, then $y$ should also be some decimal to make the equation an integer, i.e., $7$. So $y$ will be a decimal in this case.

But consider when $x^{\frac{3}{2}}$ is an integer. The only integers whose difference of squares is $7$, is $4$ and $3$ (why?).
$$4^2 - 3^2 =7$$
But compare it to the equation $y^2-(x^{\frac{3}{2}})^2=7$. Then
$$x^{\frac{3}{2}} = 3$$
$$x = 3^{\frac{2}{3}}$$
But here $x$ is a decimal!

Therefore, at least one of the variables $x$ and $y$ is a decimal in the given equation, thus proving the fact that it has no integral solutions.
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