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Old 2006-07-14, 16:58   #1
Wacky
 
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Default PR 4 # 33 -- The last puzzle from this series

Prove that the product of 4 consecutive positive integers cannot be a perfect square.
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Old 2006-07-14, 17:06   #2
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one of the numbers is divisible by 4 and one by 2. 8 is not a perfect square.

Another question. Is there a value n, n>2, such that the product of n consecutive integers is a perfect square.

Last fiddled with by Citrix on 2006-07-14 at 17:08
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Old 2006-07-14, 17:11   #3
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Quote:
Originally Posted by Citrix
one of the numbers is divisible by 4 and one by 2. 8 is not a perfect square.
What prevents the "multiple of 4" from being a multiple of 8?
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Old 2006-07-14, 17:11   #4
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Quote:
Originally Posted by Citrix
one of the numbers is divisible by 4 and one by 2. 8 is not a perfect square.

Another question. Is there a value n, n>2, such that the product of n consecutive integers is a perfect square.
Yes, and n can be arbitrarily large. Any range which includes zero satisfies the problem as stated.

Paul

Last fiddled with by xilman on 2006-07-14 at 17:11
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Old 2006-07-14, 17:16   #5
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Axn1, you are right, I did not think of that.

Xilman, if you do not include zero, what t!/(t-n)! can ever be a prefect square?
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Old 2006-07-14, 23:08   #6
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n*(n+1)*(n+2)*(n+3) = n^4 + 6*n^3 + 11*n^2 + 6*n

((n+1)*(n+2)-1)^2 = n^4 + 6*n^3 + 11*n^2 + 6*n + 1

So, we need two perfect squares with a difference of 1. 0 and 1, anyone?
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Old 2006-07-20, 14:10   #7
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Quote:
Originally Posted by Wacky
Prove that the product of 4 consecutive positive integers cannot be a perfect square.
There are two cases: 1*2*3*4 = 24 (which is not square) and the rest:

Since the difference between any member of this set is not greater than 3, the gcd of two of these numbers cannot be greater than 3.

This means that any prime factor greater than 3, can appear only in one number of the set. It is possible that this prime factor can appear twice (or even number of times) in a particular number, but since there is at most one square number inside the set, there must be a prime > 3 that appear an odd number of times in a member of the set and does not appear in the other members of the set.

This implies that the product cannot be a perfect square.

Last fiddled with by alpertron on 2006-07-20 at 14:10
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Old 2006-07-21, 01:09   #8
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alpertron:

I don't see why your approach would prevent
the four consecutive numbers from being
27*a[sup]2[/sup]
8*b[sup]2[/sup]
c[sup]2[/sup]
6*d[sup]2[/sup]
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Old 2006-07-21, 15:05   #9
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Quote:
Originally Posted by wblipp
alpertron:
I don't see why your approach would prevent
the four consecutive numbers from being
27*a2
8*b2
c2
6*d2
You are right, it does not prevent it, but your sequence cannot exist:

27 a2+1 = 8 b2
27 a2+1 = 0 (mod 8)
27 a2 = -1 (mod 8)
27 a2 = 7 (mod 8)
3 a2 = 7 (mod 8)
a2 = 5 (mod 8)

but a2 must be 0 or 1 (mod 8)

If you exchange the first and third members of the sequence you get:

27 a2-1 = 8 b2
27 a2-1 = 0 (mod 8)
27 a2 = 1 (mod 8)
3 a2 = 1 (mod 8)
a2 = 3 (mod 8)

so it is also invalid.

Since 32 = 1 (mod 8), the number 27 cannot be replaced by 27*32k.

Last fiddled with by alpertron on 2006-07-21 at 15:15
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Old 2006-07-21, 15:15   #10
mfgoode
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Thumbs down PR#25

Quote:
Originally Posted by wblipp
alpertron:

I don't see why your approach would prevent
the four consecutive numbers from being
27*a[sup]2[/sup]
8*b[sup]2[/sup]
c[sup]2[/sup]
6*d[sup]2[/sup]

:smile:
None of the above answers are convincing enough. Personally I would prefer a more rigorous proof, a real hard boiled one, with no ambiguity or speculation.

HINT: let the product be P, then in all cases it is one short of a perfect square Thus P + 1 = a perfect square and I leave it up to all to prove this

I will wait overnight and then will give a rigorous proof tomorrow.


Mally
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Old 2006-07-21, 15:33   #11
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Quote:
Originally Posted by mfgoode

:smile:
None of the above answers are convincing enough. Personally I would prefer a more rigorous proof, a real hard boiled one, with no ambiguity or speculation.

HINT: let the product be P, then in all cases it is one short of a perfect square Thus P + 1 = a perfect square and I leave it up to all to prove this

I will wait overnight and then will give a rigorous proof tomorrow.


Mally
May I refer you to Post #6?

n*(n+1)*(n+2)*(n+3) = n^4 + 6*n^3 + 11*n^2 + 6*n

((n+1)*(n+2)-1)^2 = n^4 + 6*n^3 + 11*n^2 + 6*n + 1
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