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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.

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There are $1$-$2$ additional ways of solving this here: math.stackexchange.com/questions/828024/… –  mathh Jun 11 at 10:29
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3 Answers 3

up vote 45 down vote accepted

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

That's all :)

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

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It was exactly i was writting.... –  soumajit das May 9 at 5:59
    
@soumajit das, nice. –  Oleg567 May 9 at 6:01
    
but i type slow...thats the problem.. –  soumajit das May 9 at 6:02
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I like this, thank you. Is my proof still correct though? –  Pavelshu May 9 at 6:04
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@Pavelshu Yes, your proof is correct. –  Goos May 9 at 21:29
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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: $$ n^2<n^2+n<n^2+2n+1$$

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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.

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