prove that

**archimedes81** · May 31st 2009, 01:09 PM

prove that the product of five successive natural numbers can't be a perfect square.

**CaptainBlack** · June 1st 2009, 10:08 PM

Originally Posted by archimedes81

prove that the product of five successive natural numbers can't be a perfect square.

Let the five numbers be: $n,\ n+1,\ n+2,\ n+3,\ n+4$ their product is divisible by $n$ , and if $n>4$ is not divisible by $n^2$ as that would require that $1\ 2,\ 3$ or $4$ were divisible by $n$ .

The cases $n=1,2,3,4$ can be eliminated by direct calculation.

CB

**Moo** · June 3rd 2009, 12:12 AM

Hi,

Originally Posted by CaptainBlack

Let the five numbers be: $n,\ n+1,\ n+2,\ n+3,\ n+4$ their product is divisible by $n$ , and if $n>4$ is not divisible by $n^2$ as that would require that $1\ 2,\ 3$ or $4$ were divisible by $n$ .

The cases $n=1,2,3,4$ can be eliminated by direct calculation.

CB

I had a thought on this problem... And started like you. But let's imagine n=pq.
Then, we can have p divides 2 (p=2) and q divides 4 (q=4). But yet n=8 and doesn't divide 1,2,3,4.

Whereas if we consider the consecutive numbers n-2,n-1,n,n+1,n+2, we have their product equal to :
$n(n^2-1)(n^2-4)=n(n^4-5n^2+4)$
And from this, it should be clear that n has to divide 4.

BUT there is still a problem. What happens if n is a square ?

Or if n contains a square in its factors ???
We have to reconsider the equation

**CaptainBlack** · June 3rd 2009, 02:35 AM

Originally Posted by Moo

Hi,

I had a thought on this problem... And started like you. But let's imagine n=pq.
Then, we can have p divides 2 (p=2) and q divides 4 (q=4). But yet n=8 and doesn't divide 1,2,3,4.

Consider the prime decomposition of $n=p_1^{k_1} ... p_r^{k_r}$

We have $(n+1)(n+2)(n+3)(n+4)$ is divisible by n (if the product of the five consecutive numbers is a perfect square). So $p_i$ divides one of $2,3$ or $4$ , and the only possibilities this leaves for $n$ are $2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24$ and these can be eliminated by direct computation (I could add additional justification here but the reader should be able provide that for themselves - if I have got the argument right).

That should work, or have I still missed something?

CB

**Moo** · June 3rd 2009, 02:54 AM

Originally Posted by CaptainBlack

Consider the prime decomposition of $n=p_1^{k_1} ... p_r^{k_r}$

We have $(n+1)(n+2)(n+3)(n+4)$ is divisible by n (if the product of the five consecutive numbers is a perfect square). So $p_i$ divides one of $2,3$ or $4$ , and the only possibilities this leaves for $n$ are $2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24$ and these can be eliminated by direct computation (I could add additional justification here but the reader should be able provide that for themselves - if I have got the argument right).

That should work, or have I still missed something?

CB

This works, yes ^^

But now, you haven't considered the situation where n has a square in its factor (or even if it's a square itself).
Because then we wouldn't need n to divide (n+1)(n+2)(n+3)(n+4)

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