SOLUTION: Prove that if 1 is added to the product of any four consecutive integers, the sum is a perfect square. Thank you in advance.

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Question 1059605: Prove that if 1 is added to the product of any four
consecutive integers, the sum is a perfect square.
Thank you in advance.
Found 3 solutions by ikleyn, Boreal, KMST:
Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
Prove that if 1 is added to the product of any four
consecutive integers, the sum is a perfect square.
Thank you in advance.
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The problem asks us to prove that n*(n+1)*(n+2)*(n+3) + 1 is a square of an integer.


Let x = %28n%2B1%29+%2B+1%2F2 be the central point for the original four integers n, n+1, n+2 and n+3.  Then

  n*(n+1)*(n+2)*(n+3) + 1 = (x+0.5)*(x-0.5)*(x+1.5)*(x-1.5) + 1 = 

= %28x%5E2-0.25%29%2A%28x%5E2-2.25%29+%2B+1 = x%5E4+-+2.5x%5E2+%2B+%282.25%2F4%29+%2B+1 = 

= x%5E4+-+2.5x%5E2+%2B+6.25%2F4 = %28x%5E2-2.5%2F2%29%5E2.


Next,  x%5E2+-+2.5%2F2 = (by the definition of "x") = %28n%2B1%29%5E2+%2B+%28n%2B1%29+%2B+%281%2F2%29%5E2+-+5%2F4 = %28n%2B1%29%5E2+%2B+%28n%2B1%29+-+1  is the integer number.


Thus we proved that  n*(n+1)*(n+2)*(n+3) + 1  is the square of the integer number  %28n%2B1%29%5E2+%2B+%28n%2B1%29+-+1 = %28n%2B1%29%5E2+%2B+n.

Proved and solved.


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x(x+1)(x+2)(x+3)+1=x(x+3)(x+2)(x+1)+1
so (x^2+3x)(x^2+3x+2)+1=(x^2+3x+1)(x^2+3x+2)-(x^2+3x+2)+1
You are converting the first term to (x^2+3x+1), and that is adding x^2+3x+2 to that side. That means you have to subtract it.
Convert the (x^2+3x+2) to (x^2+3x+1). But that requires adding the factor x^2+3x+1. because by subtracting one from that term, which is multiplied by (x^2+3x+1), you have to add it back.
Then you have (x^2+3x+1)(x^2+3x+1)-(x^2+3x+2)+(x^2+3x+1)+1. The last two terms disappear because distributing the minus sign give -x^2-3x-2+x^2+3x+1+1.
You are left with (x^2+3x+1)^2, which is a perfect square.
Pick an x, like 7
7,8,9,10 has a product of 5040 and 5041, 1 more, is a perfect square of
x^2+3x+1, or 49+21+1, or 71.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the four consecutive integers be defined based on their avetshe, a , as
a-3%2F2 , a-1%2F2 , a%2B2%2F2 , and a%2B3%2F2 .
The sum the problem talks about is
%28a-3%2F2%29%28a-1%2F2%29%28a%2B1%2F2%29%28a%2B3%2F2%29%2B1=
%28a%5E2-9%2F4%29%28a%5E2-1%2F4%29%2B1=
a%5E4-%2810%2F4%29a%5E2%2B9%2F16%2B1=
a%5E4-2%285%2F4%29a%5E2%2B25%2F16=
%28a%5E2-5%2F4%29%5E2 .
a%5E2-5%2F4 is an integer.
In fact, it is
the sum of 1 plus the product if the first and fourth integers:
1%2B%28a-3%2F2%29%28a%2B3%2F2%29=4%2F4%2Ba%5E2-9%2F4=a%5E2-5%2F4 .