20060714, 16:58  #1 
Jun 2003
The Texas Hill Country
1089_{10} Posts 
PR 4 # 33  The last puzzle from this series
Prove that the product of 4 consecutive positive integers cannot be a perfect square.

20060714, 17:06  #2 
Jun 2003
11^{2}·13 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 20060714 at 17:08 
20060714, 17:11  #3  
Jun 2003
37·127 Posts 
Quote:


20060714, 17:11  #4  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
5·2,053 Posts 
Quote:
Paul Last fiddled with by xilman on 20060714 at 17:11 

20060714, 17:16  #5 
Jun 2003
11^{2}·13 Posts 
Axn1, you are right, I did not think of that.
Xilman, if you do not include zero, what t!/(tn)! can ever be a prefect square? 
20060714, 23:08  #6 
Jun 2003
1001001011011_{2} 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? 
20060720, 14:10  #7  
Aug 2002
Buenos Aires, Argentina
52D_{16} Posts 
Quote:
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 20060720 at 14:10 

20060721, 01:09  #8 
"William"
May 2003
New Haven
4466_{8} 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] 
20060721, 15:05  #9  
Aug 2002
Buenos Aires, Argentina
10100101101_{2} Posts 
Quote:
27 a^{2}+1 = 8 b^{2} 27 a^{2}+1 = 0 (mod 8) 27 a^{2} = 1 (mod 8) 27 a^{2} = 7 (mod 8) 3 a^{2} = 7 (mod 8) a^{2} = 5 (mod 8) but a^{2} must be 0 or 1 (mod 8) If you exchange the first and third members of the sequence you get: 27 a^{2}1 = 8 b^{2} 27 a^{2}1 = 0 (mod 8) 27 a^{2} = 1 (mod 8) 3 a^{2} = 1 (mod 8) a^{2} = 3 (mod 8) so it is also invalid. Since 3^{2} = 1 (mod 8), the number 27 cannot be replaced by 27*3^{2k}. Last fiddled with by alpertron on 20060721 at 15:15 

20060721, 15:15  #10  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
PR#25
Quote:
: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 

20060721, 15:33  #11  
Aug 2005
Brazil
2·181 Posts 
Quote:
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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