$ (x-2)(x-1)x(x+1)(x+2)=a^2\Longrightarrow x^2-1=u@,x^2-4=v^2$ or $ x^2-1=3u^2,x^2-4=3v^2$ - had not solution.

A typical terse, and incomplete "solution" from Rust. We want $ (x-2)(x-1)x(x+1)(x+2) = x(x^2-1)(x^2-4) = a^2$. We have $\gcd(x,x^2-1) = 1$, $\gcd(x,x^2-4) \mid 4$, $\gcd(x^2-1,x^2-4) \mid 3$. If $\gcd(x,x^2-4) =1$ and $\gcd(x^2-1,x^2-4) = 1$, it follows $x^2-1$, $x^2-4$ must be perfect squares, impossible. If $\gcd(x,x^2-4) =1$ and $\gcd(x^2-1,x^2-4) = 3$, it follows $x\not \equiv 0 \pmod{3}$, and so $3(x^2-1)$, $3(x^2-4)$ must be perfect squares, impossible. If $\gcd(x,x^2-4) > 1$ it follows $x=2y$, and so $x^2-4 = 4(y^2-1)$. Since then $x^2-1$ is odd, it follows we need have $y=2z$, and so $x = 4z$, $x^2-4 = 4(4z^2-1)$. If now $\gcd(x^2-1,x^2-4) = 1$, it follows $z$, $4z^2-1$ must be perfect squares, impossible. Finally, if $\gcd(x^2-1,x^2-4) = 3$, another analysis must be done, but I run out of patience. Of course, by a result of Erdös and Selfridge, no product of consecutive positive integers is ever a perfect power.

Whoever is Reading this Post, I advise that I am usually online from Monday to Friday at 6: 30. Now, this problem is a Good introdution to study the Erdos Selfridge theorem. One solution basically consists in writing each of the five numbers as the product of a square and a square free integer, and notice that there are only 4 possibilities for the values of the square free numbers; since we have five numbers, by the Pigeonhole principle, two share the same "square free part". To finish the problem, it is suffice to consider each of the 4 cases and note that the diference of two numbers is bounded above by 4; we find that there is only one case to consider: squarefree part equal 1; and numbers 1 and 4, i.e., the case 5!, which is not a square.

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