Hi Liza,
We have two responses for you, but before posting them I want to summarize what they did as I find it quite instructive.
Sue played with your observation about the product of 4 consecutive positive integers and conjectured
k(k+1)(k+2)(k+3) + 1 = ((k+2)(k+1)  1)^{2}
She then proved this two ways, once by expanding both sides and verifying they are equal and again by factoring the left side to arrive at the right side. Both of these arguments required some careful and involved algebra.
Penny wrote the product of 4 consecutive positive integers as
(n2)(n1)n(n+1)
She didn't include the details that Sue included but I find it instructive that this simple shift in the variable made the algebraic expressions look considerably simpler.
Harley
Hi Liza. That is an interesting property.
I tried the next couple of products:
3x4x5x6 + 1 = 19^2^{2}
and
4x5x6x7 + 1 = 29^{2}
Looking at this sequence of squares...
start with 1, get 5^{2}
start with 2, get 11^{2}
start with 3, get 19^{2}
start with 4, get 29^{2}
...there's a pattern I see:
5 = 3x2  1
11 = 4x3  1
19 = 5x4  1
29 = 6x5  1
It looks to me as though if you start with k,
you get the square of (k+2)(k+1)  1.
Now we have to prove it.
Let k be some integer greater than zero. Then we want to show that
k(k+1)(k+2)(k+3) + 1 = ((k+2)(k+1)  1)^{2}
Since factoring quartics is not something I find particularly easy,
I'll just multiply it out.
Left hand side:
k(k+1)(k+2)(k+3) + 1
= (k^{2} + k)(k^{2} + 5k + 6) + 1
= k^{4} + 5k^{3} + 6k^{2} + k^{3} + 5k^{2} + 6k + 1
= k^{4} + 6k^{3} + 11k^{2} + 6k + 1.
Right hand side:
((k+2)(k+1)  1)^{2}
= (k^{2} + 3k + 2  1)^{2}
= (k^{2} + 3k + 1)^{2}
= k^{4} + 3k^{3} + k^{2}2 + 3k^3 + 9k^{2} + 3k + k^{2} + 3k + 1
= k^{4} + 6k^{3} + 11k^{2} + 6k + 1
They match, so this works. I hope this helps you see what is
happening, even though the algebra is a bit more involved than
usual.
Stephen La Rocque.PS: I tried it by factoring after all, Liza. Here's what happened:
k(k+1)(k+2)(k+3) + 1
= (k+1)(k+2)(k^{2}+3k) + 1
= ((k+1)(k+2)  1) (k^{2}+3k) + k^{2} + 3k + 1
= ((k+1)(k+2)  1) (k^{2}+3k) + k^{2} + 3k + 2  1
= ((k+1)(k+2)  1) (k^{2}+3k) + ((k+2)(k1)  1)
= ((k+1)(k+2)  1) ((k^{2}+3k) + 1 )
= ((k+1)(k+2)  1) ((k^{2}+3k+2)  1 )
= ((k+1)(k+2)  1) ((k+1)(k+2)  1)
= ((k+1)(k+2)  1)^{2}
QED
Hi Liza,
The pattern of numbers is
(n2)x(n1)x(n)x(n+1) + 1 = (n^{2}2n)(n^{2}1) +1 = n^{4}  2n^{3}  n^{2} + 2n + 1 = (n^{2}n1)^{2}.
Penny
