The product of 8 consecutive natural numbers will not be a perfect square?

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I wish to show that the sum of 9 consecutive natural numbers: n(n+1)(n+2)(n+3)...(n+8) will not be a perfect square.
This problem came by as a result of another problem I was doing and I'm wondering if anyone knows/has come across this already.

After some searching I found that n! is not a perfect square (although I found no proof).

In my case I wish to show that (n+8)!-(n-1)! is not a perfect square.
 

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  • #2
haruspex
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I wish to show that the sum of 9 consecutive natural numbers: n(n+1)(n+2)(n+3)...(n+8) will not be a perfect square.
This problem came by as a result of another problem I was doing and I'm wondering if anyone knows/has come across this already.

After some searching I found that n! is not a perfect square (although I found no proof).

In my case I wish to show that (n+8)!-(n-1)! is not a perfect square.
Your title says product of 8, your opening text says sum of 9, which you then illustrate with a product of nine. And you finish with an expression which is the difference of two factorials (I guess you meant the ratio?). Please clarify which case you're interested in.

As for n! never being a square, can you show that there is always a prime between n and n2?
 
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Oh sorry about the confusion...i meant consecutive 9.. as for the difference I think it is correct...the main problem is to show n(n+1)..(n+8) is not a perfect square...
As for n!..i did a bit more searching and found bertrands postulate to show there is such a prime..i should be able to figure it out from there ( hopefully)
 
  • #4
haruspex
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as for the difference I think it is correct...the main problem is to show n(n+1)..(n+8) is not a perfect square...
(n+8)! - (n-1)! is very different from n(n+1)..(n+8). You mean (n+8)! / (n-1)!
As for n!..i did a bit more searching and found bertrands postulate to show there is such a prime..i should be able to figure it out from there ( hopefully)
Sorry, I should have said "between n and 2n".
 
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Oh wow..you're right about the division..that was silly
 
  • #6
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Okay how about this:
We wish to prove n(n+1)(n+2)....(n+m) is not a perfect square. Assume to the contrary that it is a perfect square. Then, the prime factors can be written to an even power. However, this is a contradiction because there exists a largest prime (by bertrands postulate).
 
  • #7
haruspex
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Okay how about this:
We wish to prove n(n+1)(n+2)....(n+m) is not a perfect square. Assume to the contrary that it is a perfect square. Then, the prime factors can be written to an even power. However, this is a contradiction because there exists a largest prime (by bertrands postulate).
That's a dizzying leap of logic in the last line.
Does it work with m=0:wink:?
Consider also 8*9. All primes in there have powers > 1, it's just that one or more is odd. So I doubt the proof will be simple.
 
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right...m>0...and your example just wrecked my proof completely :)... here's another try:

Assume G=n(n+1)(n+2)...(n+m) is a perfect square (m>0,n>0). Consider the set of prime factors of G: {p1,p2,....pn} where pn is less than or equal to m because if it were greater than m, G would have at most 1 factor of pn. Note that 2|G. Now we consider the repeated powers of each prime: 4 can divide G at most ....

This is where I get stuck. I would like to show the maximum number of times that pn^2 divides g (for each n) and then select from the remaining non repeating powers and conclude that the sequence contradicts the assumption of a prime larger than pn (the last n)....
 
  • #9
haruspex
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Assume G=n(n+1)(n+2)...(n+m) is a perfect square (m>0,n>0). Consider the set of prime factors of G: {p1,p2,....pn} where pn is less than or equal to m because if it were greater than m, G would have at most 1 factor of pn.
No. There would only be one number in there that's divisible by pn, but it might be divisible an even number of times.
Note that 2|G. Now we consider the repeated powers of each prime: 4 can divide G at most ....

This is where I get stuck. I would like to show the maximum number of times that pn^2 divides g (for each n) and then select from the remaining non repeating powers and conclude that the sequence contradicts the assumption of a prime larger than pn (the last n)....
I think you'll find that for each prime factor of G there's no way to exclude the possibility that is has an even exponent in G. What seems to happen is that they can't all have even exponents.
 

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