<s>[INST] Find all real $x$ ,such $8x^3-20$ and $2x^5-2$ is perfect square 
Find all real numbers $x$,such 
  $$8x^3-20,2x^5-2$$  is  the perfect square of an integer

My idea: First we find the real number $x$ such $$8x^3-20,2x^5-2$$ is postive integer numbers,and second the real numbers such
$$8x^3-20=m^2,2x^5-2=n^5$$
How prove this all real numbers?
 [/INST] First observe that $x$ is a positive integer. $x^5$ and $x^3$ are rationals; then $x=(x^3)^2/x^5$ is rational; finally $(2x)^5$ is an even positive integer so $2x$ must be an even positive integer. 
In order to find a bound for $x$, consider the product of $8x^3−20=m^2$ and $2x^5−2=n^2$ and complete the square.
\begin{align*}
(mn)^2 &= (8x^3-20)(2x^5-2) \\ 
&= 16x^8 - 40x^5 - 16x^3 + 40 \\
&= (4x^4-5x)^2 -(16x^3+25x^2-40) \\
&= (4x^4-5x-1)^2 +(8x^4-16x^3-25x^2-10x+39).
\end{align*}
If $x\ge4$ then the remaining terms
$16x^3+25x^2-40$ and $8x^4-16x^3-25x^2-10x+39$ are positive and we find
$$
4x^4-5x-1 < mn < 4x^4-5x
$$
that is impossible. Hence, $x\le3$.
For $x=1$ we have $8x^3-20<0$, no solution.
For $x=2$ we have $8x^3-20=44$, no solution.
For $x=3$ we have $8x^3-20=14^2$ and $2x^5-2=22^2$.
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