1. ## prove that

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

2. Originally Posted by archimedes81
prove that the product of five successive natural numbers can't be a perfect square.
Let the five numbers be: $\displaystyle n,\ n+1,\ n+2,\ n+3,\ n+4$ their product is divisible by $\displaystyle n$, and if $\displaystyle n>4$ is not divisible by $\displaystyle n^2$ as that would require that $\displaystyle 1\ 2,\ 3$ or $\displaystyle 4$ were divisible by $\displaystyle n$.

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

CB

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

The cases $\displaystyle n=1,2,3,4$ 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 :
$\displaystyle n(n^2-1)(n^2-4)=n(n^4-5n^2+4)$
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

4. 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 $\displaystyle n=p_1^{k_1} ... p_r^{k_r}$

We have $\displaystyle (n+1)(n+2)(n+3)(n+4)$ is divisible by n (if the product of the five consecutive numbers is a perfect square). So $\displaystyle p_i$ divides one of $\displaystyle 2,3$ or $\displaystyle 4$, and the only possibilities this leaves for $\displaystyle n$ are $\displaystyle 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24$ 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

5. Originally Posted by CaptainBlack
Consider the prime decomposition of $\displaystyle n=p_1^{k_1} ... p_r^{k_r}$

We have $\displaystyle (n+1)(n+2)(n+3)(n+4)$ is divisible by n (if the product of the five consecutive numbers is a perfect square). So $\displaystyle p_i$ divides one of $\displaystyle 2,3$ or $\displaystyle 4$, and the only possibilities this leaves for $\displaystyle n$ are $\displaystyle 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24$ 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)