Proof: The product of any 4 consecutive integers

I was trying to manipulate it to fit the n2 + 3n + 1 pattern. In summary, the product of any 4 consecutive integers, (n)(n+1)(n+2)(n+3), can be written as ((n+1)(n+2))((n)(n+3)) which is equivalent to (n^2+3n+1)^2-1. This proves that the product of 4 consecutive integers is always one less than a perfect square.
  • #1
Animuo
13
0

Homework Statement


The product of any 4 consecutive integers will be one less than a perfect square.


Homework Equations


Well, a perfect square is a number that can be broken down to n*n where n is an integer.
If a number is consecutive to another number that means it is exactly one more than the other.


The Attempt at a Solution


So you substitute n for "any integer" and you get this

(n) (n+1) (n+2) (n+3)

I wasn't sure where to go from here I tried making it into a big polynomial and just got confused. The basic idea is that you want to somehow group these numbers together to get two equivalent parts (contained in parentheses) minus one. I've tried several different ways of doing this but for some reason I can't figure it out... I'll let you guys do the work :), thanks ahead of time
 
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  • #2
Animuo said:
I'll let you guys do the work :)

Aww shucks :!) But we're not allowed! The big bad boss would strip me of my homework helpers medal if I did that... And then I'd have no way to feed my family! :-p

If [itex]f(n)=n(n+1)(n+2)(n+3)[/itex] is equivalent to the form of [itex]k^2-1[/itex] where k is some integer, then after expanding:

[tex]f(n)=n^4+6n^3+11n^2+6n[/tex] we want to get it in the form [itex]k^2-1[/itex] and then test to see if we can find such a k. Let's assume for a second that k exists and it's an integer, how can we manipulate f(n) to get it in such a form? Notice f(n) has no constants, which we'll need.
 
  • #3
I would add 1 to that "big polynomial" and try to see if I could factor it as a perfect square. Maple will do it for you but I suppose that defeats the point of the problem.
 
  • #4
Uggh I am literally soo bad at factoring.. which is why I kinda handed the reins over.. anyways by trial and error, and some common-sense I came up with (n^2 + 3n + 1)^2 = k^2, so that proves the four consecutive numbers produce a perfect square minus one. I had a lot of trouble factoring this though, and I think that was biggest problem, high school's becoming a distant memory for me, and the math I'm teaching myself now expects me to be pretty fluent at it.. could someone shoot me a link explaining how to factor larger polynomials, or maybe quickly explain it.. thanks for the help guys, it pushed me in the right direction
 
  • #5
I will confess that my first thought was "that can't possibly be true" so I started looking for a counter example: [itex]1(2)(3)(4)= 24= 5^2- 1[/itex], [itex]2(3)(4)(5)= 120= 11^2- 1[/itex], [itex]3(4)(5)(6)= 360= 19^2- 1[/itex], [itex]4(5)(6)(7)= 840= 29^2- 1[/itex], [itex]5(6)(7)(8)= 1680= 41^2- 1[/itex].
Well, gosh, maybe it is true! Further, that list shows that the product of four integers, n(n+1)(n+3)(n+4) is [itex]k^2- 1[/itex] so that I can associate each n with a corresponding k: 1:5, 2:11, 3:19, 4:29, 5:41. And now I could use Newton's divided difference method to find that [itex]k= n^2+ 3n+ 1[/itex], at least for those numbers.

But knowing that, I can calculate both [itex](n^2+ 3n+ 1)^2- 1[/itex] and [itex]n(n+1)(n+3)(n+ 4)[/itex] and see that they are the same for all n.
 
  • #6
Animuo said:
Uggh I am literally soo bad at factoring.. which is why I kinda handed the reins over.. anyways by trial and error, and some common-sense I came up with (n^2 + 3n + 1)^2 = k^2, so that proves the four consecutive numbers produce a perfect square minus one. I had a lot of trouble factoring this though, and I think that was biggest problem, high school's becoming a distant memory for me, and the math I'm teaching myself now expects me to be pretty fluent at it.. could someone shoot me a link explaining how to factor larger polynomials, or maybe quickly explain it.. thanks for the help guys, it pushed me in the right direction

I don't recall ever reading much about how to factor higher degree polynomials that aren't some special case. But after knowing the answer from Maple and finding the answer was a trinomial squared (I know, I know) I wondered if using that I could factor$$
n^4 +6n^3 + 11n^2 + 6n + 1$$Looking at the highest and lowest order terms it seemed clear the first and last terms of the trinomial must be ##n^2## and ##1##. Now$$
(n^2 + f(n) + 1)^2 = n^4+2n^2f(n) + 2n^2 +f^2(n) + 2f + 1$$That takes care of the first and last terms and leaves us with$$
2n^2f(n) + 2n^2 +f^2(n) + 2f=6n^3 + 11n^2 + 6n$$ Then by inspection I observed ##f(n) = 3n## gives the first and last terms here and, luckily, the middle two terms give the ##11n^2##. So we get ##(n^2 + 3n + 1)^2##.
 
  • #7
Animuo said:

Homework Statement


The product of any 4 consecutive integers will be one less than a perfect square.


Homework Equations


Well, a perfect square is a number that can be broken down to n*n where n is an integer.
If a number is consecutive to another number that means it is exactly one more than the other.

The Attempt at a Solution


So you substitute n for "any integer" and you get this

(n) (n+1) (n+2) (n+3)

I wasn't sure where to go from here I tried making it into a big polynomial and just got confused. The basic idea is that you want to somehow group these numbers together to get two equivalent parts (contained in parentheses) minus one. I've tried several different ways of doing this but for some reason I can't figure it out... I'll let you guys do the work :), thanks ahead of time
You can write [itex]\displaystyle (n) (n+1) (n+2) (n+3)[/itex] as

[itex]\displaystyle ((n+1)(n+2))\, ((n)(n+3))[/itex]
[itex]\displaystyle= (n^2+3n+2)\, (n^2+3n)[/itex]

[itex]\displaystyle =((n^2+3n+1)+1)\, ((n^2+3n+1)-1)[/itex]  ... (a sum times a difference which is the difference of squares)

[itex]\displaystyle =(n^2+3n+1)^2-1[/itex]
 
  • #8
LCKurtz said:
$$
n^4 +6n^3 + 11n^2 + 6n + 1$$Looking at the highest and lowest order terms it seemed clear the first and last terms of the trinomial must be ##n^2## and ##1##.

I went with that idea that n2 and 1 were the first and last terms of the trinomial. Then I wrote it as (n4 + 6n3 + n2) + (n2 + 6n + 1) + 9n2 = n2(n2 + 6n + 1) + (n2 + 6n + 1) + 9n2 so the stuff in parentheses is the same, but a dead end since that 9n2 is all by itself.
So then I first broke up the terms with 6 as 3+3 and brought one of each of them over with the 9 term to see if they could factor nicely:

n4 + 6n3 + 11n2 + 6n + 1

= n4 + 3n3 + n2 + n2 + 3n + 1 + 9n2 + 3n2 + 3n

= n2(n2 + 3n + 1) + (n2 + 3n + 1) + 3n(n2 + 3n + 1) = (n2 + 3n + 1)(n2 + 3n + 1) = (n2 + 3n + 1)2
 
  • #9
Thanks guys, I already figured out the solution but some of you guys employ some interesting methodology.. I'll give it a look when I have more time and see what I think.
 
  • #10
SammyS said:
You can write [itex]\displaystyle (n) (n+1) (n+2) (n+3)[/itex] as

[itex]\displaystyle ((n+1)(n+2))\, ((n)(n+3))[/itex]
[itex]\displaystyle= (n^2+3n+2)\, (n^2+3n)[/itex]

[itex]\displaystyle =((n^2+3n+1)+1)\, ((n^2+3n+1)-1)[/itex]  ... (a sum times a difference which is the difference of squares)

[itex]\displaystyle =(n^2+3n+1)^2-1[/itex]

Ahh that's neat :smile:
 

FAQ: Proof: The product of any 4 consecutive integers

What is the product of any 4 consecutive integers?

The product of any 4 consecutive integers is equal to the first integer multiplied by the third integer plus one.

Can this proof be applied to any set of consecutive integers?

Yes, this proof can be applied to any set of consecutive integers, as long as there are at least 4 integers in the set.

How can this proof be used in real-world applications?

This proof can be used in various mathematical and scientific calculations, such as finding the volume of a rectangular prism or calculating the distance of an object in motion based on its velocity and time.

Is this proof considered a mathematical law or theory?

This proof is considered a mathematical law, as it has been proven to be true for all sets of consecutive integers.

Can this proof be extended to more than 4 consecutive integers?

Yes, this proof can be extended to any number of consecutive integers by using the same formula of the first integer multiplied by the last integer plus one.

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