Show that the product of two consecutive natural numbers is never a square.

Question

I'd like to have my proof verified and if possible, to see other solutions that are interesting.

Proof: Suppose $n(n+1)$ is a square. Then we write $$n(n+1) = \prod_{p} p^{c(p)}$$ where $c(p) = a(p) + b(p)$ are such that \begin{align*} n &= \prod_p p^{a(p)} \\ n+1 &= \prod_p p^{b(p)} \end{align*} Now, by our hypothesis, $c(p)$ is even for all primes $p$. As $(n,n+1)=1$ for all $n$, it must be that $a(p)$ and $b(p)$ are even for all primes $p$ and moreover, $a(p) = 0$ whenever $b(p)>0$ and reversely.

This indicates that both $n$ and $n+1$ are squares. This is impossible as there are no consecutive squares in the natural numbers.

There are $1$-$2$ additional ways of solving this here: http://math.stackexchange.com/questions/828024/how-to-prove-that-nn1-cant-be-square/828102#comment1711473_828102 — user26486, Jun 11 '14 at 10:29

score 57 · Accepted Answer · answered May 09 '14 at 05:59

57

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

That's all :)

There are no integer number between $n$ and $n+1$.

answered May 09 '14 at 05:59

Oleg567

17,295

1

It was exactly i was writting.... – soumajit das May 09 '14 at 05:59
@soumajit das, nice. – Oleg567 May 09 '14 at 06:01
but i type slow...thats the problem.. – soumajit das May 09 '14 at 06:02
1

I like this, thank you. Is my proof still correct though? – Pavelshu May 09 '14 at 06:04
@soumajit das, you'll be first in other time. anyway. Sometimes this is like sport. – Oleg567 May 09 '14 at 06:04
@ Pavelshu..it is correct but why go for such rigorous method.... – soumajit das May 09 '14 at 06:06
@Pavelshu, I still didn't read your proof carefully. – Oleg567 May 09 '14 at 06:06
2

@Pavelshu Yes, your proof is correct. – Caleb Stanford May 09 '14 at 21:29

score 11 · Answer 2 · answered May 09 '14 at 05:59

11

This is fine if $n=0$ is not allowed (and if $n=0$ is allowed the claim becomes wrong. As you ask for alternative proofs, what can you see from this inequality: $$ n^2<n^2+n<n^2+2n+1$$

answered May 09 '14 at 05:59

Hagen von Eitzen

374,180

score 5 · Answer 3 · answered May 09 '14 at 06:07

It looks like this may be the same proof as the OP, but here's how I'd write it:

Let $n$ and $n+1$ be consecutive natural numbers. Note that they must have disjoint prime factorizations. I.e. if $p|n$, then $p \nmid (n+1)$.

For $n(n+1)$ to be a perfect square, then the power on every prime in its decomposition must be even. However, since $n$ and $n+1$ have disjoint prime decompositions, then this is only possible if the power on every prime in the decomposition of $n$ is even, and similarly for $n+1$. This would imply that both $n$ and $n+1$ are perfect squares, which is impossible since no two consecutive naturals can be perfect squares.

Granted, this proof is a lot more clumsy than Hagen's and Oleg's.

Farbod Ahmadian · Answer 4 · 2019-08-26T11:44:12.920

Well I have another explain for it, but I'm not sure if it's true or not, So I will be happy that somebody correct me :)

We want to discuss about if n(n+1) is square or not, so we try to make it square:

$n(n+1) = n^2 + n = n^2 + n + \frac 14 - \frac 14 = (n+\frac12)^2 - \frac 14$

Now let X = $(n+\frac12)$

Now we have:

$$ n(n+1) = X^2 - \frac 14$$

So as you can see, this equation shows us that $n(n+1)$ is always not equal to a square.

Show that the product of two consecutive natural numbers is never a square.

4 Answers4

Linked

Related