<s>[INST] A non-inductive proof regarding a property of natural numbers Here is the question:

For each natural number $n$, prove that
$$n(n + 1)(n +2) (n+3)$$
is never a perfect square.

My approach: Here,we notice that two of the factors given in the product are bound to be odd while
the remaining two are bound to be even. WLOG,let $n$ be even. If $n=2k$ for some integer $k$ then
$$n+2=2k+2=2(k+1).....(1)$$. Thus,we get that $n(n+2)=4(k+1)k$ where $k$ and $k+1$ can have no common factor.
If two consecutive odds $n+1$ and $n+3$ have a common factor $x$ then,
$$n+3-(n+1)=2=x(k_1-k_2)$$where $n+3=k_1x$ and $n+1=k_2x$. But,then $x=1$ is the only possible value. Thus, we get that their gcd is 1.Since,we now see that there can be no double repetition of the prime factors except that of $4$ in:
$$n(n + 1)(n +2) (n+3)=4(k+1)k(n+3)(n+1)......(2)$$ since the consecutives cannot contribute a common factor and
the numbers that have the same kind of divisibility by $2$ cannot have common factors other than $2$ (for the even case).So,we get that the product is never a perfect square.(We see that steps (2) and (1) are invertible if $n$ is odd.
It is a bit rough and I am not sure whether this is correct.Please help me verify this and correct the proof. I want to see the defects of my proof.
 [/INST] $$(n^2+3n)^2<n(n+1)(n+2)(n+3)=n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2)=(n^2+3n)^2+2(n^2+3n)<(n^2+3n)^2+2(n^2+3n)+1=(n^2+3n+1)^2$$
$n(n+1)(n+2)(n+3)$ is between two consecutive perfect squares, it is not a perfect square
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