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 2006-07-14, 16:58 #1 Wacky     Jun 2003 The Texas Hill Country 32·112 Posts PR 4 # 33 -- The last puzzle from this series Prove that the product of 4 consecutive positive integers cannot be a perfect square.
 2006-07-14, 17:06 #2 Citrix     Jun 2003 30648 Posts 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
2006-07-14, 17:11   #3
axn

Jun 2003

537710 Posts

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?

2006-07-14, 17:11   #4
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

59·193 Posts

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

 2006-07-14, 17:16 #5 Citrix     Jun 2003 22×397 Posts 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?
 2006-07-14, 23:08 #6 axn     Jun 2003 19×283 Posts 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?
2006-07-20, 14:10   #7
alpertron

Aug 2002
Buenos Aires, Argentina

2×727 Posts

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

 2006-07-21, 01:09 #8 wblipp     "William" May 2003 New Haven 2,371 Posts 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]
2006-07-21, 15:05   #9
alpertron

Aug 2002
Buenos Aires, Argentina

2·727 Posts

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

2006-07-21, 15:15   #10
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
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

2006-07-21, 15:33   #11
fetofs

Aug 2005
Brazil

36210 Posts

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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