<s>[INST] Prove That $n(n+1)(n+2)(n+3)$ is never a square of a positive integer. $n \in \mathbb{N}$, bigger than zero.
My proof:(A Proof by Contradiction)
Alert: There is a lot of variables.
Suppose that:
$n(n+1)(n+2)(n+3)=x^{2}$, with $x \in \mathbb{N}$
$$n(n+3)(n+2)(n+1)=x^{2}$$
$$(n^{2}+3n)(n^{2}+3n+2)=x^{2}$$
let $n^{2}+3n = a$.
$$a(a+2)=x^{2} \iff a^{2} +2a=x^{2} \iff a^{2} +2a-x^{2}=0$$
And this is a quadratic equation in $a$, So:
$$a=\frac{-2 \pm \sqrt{4+4x^{2}}}{2} \implies \sqrt{4+4x^{2}} \in \mathbb{N} \iff 4+4x^{2}=b^{2}$$
$$ 4(1+x^{2})=b^{2} \iff 4 \mid b^{2} \iff 2 \mid b \iff b=2c.$$
Now let's plug this result to this equation:
$$4(1+x^{2})=(2c)^{2} \iff 4(1+x^{2})=4c^{2} \iff 1+x^{2}=c^{2} \iff (c-x)(c+x)=1$$
And $1=1 \cdot 1$, so :$(c-x)=1$ and $(c+x)=1$
$$c+x=c-x \iff x=-x \iff 2x=0 \iff x=0$$
By plugging this result into this equation : $a(a+2)=x^{2}.$
$a(a+2)=0 \iff a=0$ or $a=-2$ but $-2 \notin \mathbb{N}$ So : $x=0$
And finally:
$$n(n+1)(n+2)(n+3)=0 \iff n=0$$
And the other values $\notin \mathbb{N}$
This proof is too huge for me so perhaps I've made some mistakes, And if you have an easier proof please post it.
 [/INST] Hint: Show that $ n(n+1)(n+2)(n+3) + 1$ is a perfect square.
Hence, the original expression cannot be the square of a positive integer (but could be the perfect square 0).
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