A non-inductive proof regarding a property of natural numbers

Question

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.

I guess if $n$ is divisible by $3$ then, $n$ and $n+3$ can have a common factor of $3$.But,then too we are left with $9$ as the only square factor in their product. — Edmund Blackadder, Jul 17 '21 at 12:39
"WLOG, let $n$ be even": this assumption is invalid. It might be that there are solutions for odd $n$, but no solutions for even $n$. — TonyK, Jul 17 '21 at 12:52
@TonyK the working applies for any two consecutive odds and evens — Edmund Blackadder, Jul 17 '21 at 12:57
Does this answer your question? Proving any product of four consecutive integers is one less than a perfect square — lulu, Jul 17 '21 at 13:00
I guess the op is asking for a proof verification rather than an answer — snehamayee sahu, Jul 17 '21 at 13:01
It's not, or at least it is not complete. You provide no reason to imagine that it suffices to take $n$ even. Nor do you explain how the failure to repeat a prime factor $>2$ means the product is not square. — lulu, Jul 17 '21 at 13:08

score 4 · Answer 1 · answered Jul 17 '21 at 12:40

4

With $m:=n^2+3n+1$, we have $n(n+3)=n^2+3n=m-1$ and $(n+1)(n+2)=n^2+3n+2=m+1$ so that $$(m-1)^2<(m-1)(m+1)=m^2-1<m^2 .$$

answered Jul 17 '21 at 12:40

Hagen von Eitzen

374,180

score 3 · Answer 2 · answered Jul 17 '21 at 12:41

3

$$(n^2+3n)^2<n(n+1)(n+2)(n+3)=n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2)=(n^2+3n)^2+2(n^2+3n)<(n^2+3n)^2+2(n^2+3n)+1=(n^2+3n+1)^2$$

$n(n+1)(n+2)(n+3)$ is between two consecutive perfect squares, it is not a perfect square

answered Jul 17 '21 at 12:41

Lion Heart

7,073

7

I think you could go faster with $n(n+1)(n+2)(n+3)=(n^2+3n+1)^2-1$ and since no two squares can be $1$ apart (except $0^2$ and $1^2$) it is done. – zwim Jul 17 '21 at 13:02

A non-inductive proof regarding a property of natural numbers

2 Answers2