Prove that the product of three consecutive positive integers is never a perfect square.
Problem
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Tags: number theory
Honestly
08.04.2021 06:06
Assume that the product of $n-1$, $n$, and $n+1$ is a perfect square. Then, $$(n-1)(n)(n+1)=(n^2-1)(n)=k^2$$for some positive integer $k$. Note that $2|v_p(k^2)$. This is only true if both $n^2-1$ and $n$ are perfect squares as $\text{gcd}(n^2-1,n)=1$. However, $n^2-1$ cannot be a perfect square, a contradiction.
JustinLee2017
08.04.2021 06:07
Assume that the product is a perfect square. Our numbers are $(x-1), (x), $ and $(x+1)$ where $x \geq 2$. Then, we can write $$(x-1)(x)(x+1) = k^2$$for some positive integer $k$. But since $\gcd(x-1, x) = \gcd(x,x+1) = 1$, we must have $n$ being a perfect square. Then $$(x+1)(x-1) = x^2 - 1$$must also be a square. This is only true for $x = 1$, but $0$ isn't a positive integer. Thus, the product of $3$ consecutive positive integers is never a perfect square $\blacksquare$