• Support PF! Buy your school textbooks, materials and every day products Here!

Prove that the product of 4 consecutive numbers cannot be a perfect square

  • #1
139
2

Homework Statement



n(n+1)(n+2)(n+3) cannot be a square

Homework Equations



Uniqueness of prime factors for a given number

The Attempt at a Solution



I'm not sure but I think I've proved a stronger case for how product of consecutive numbers cannot be squares. I dunno whether it is right

proof :

The product of two consecutive numbers are coprime so cannot be square

in the product of three consecutive numbers eg n(n+1)(n+2), two of those consecutive numbers are divisible by 2 and not a square number but the other isn't divisible by 2 . If that other is a perfect square, then also the product of a square and a nonsquare number cannot be square .Thus 3 consecutive numbers cannot form a perfect square

Now consider 4 numbers n(n+1)(n+2)(n+4). There are 3 consecutive numbers such that their products are divisible by 3 but are not a perfect square . The other number isn't divisible by 3 therefore product of 4 consecutive numbers cannot be perfect square . Thus so on and forever.
 

Answers and Replies

  • #2
139
2
Agh, I meant product !
 
  • #3
DrClaude
Mentor
7,155
3,294
Agh, I meant product !
I corrected the title.
 
  • #4
139
2
I corrected the title.
Thanks

btw would you know why I've got no response yet? Is what I've written badly wrong or am I just not clear enough here ?
 
  • #5
33,271
4,975
Here's a different approach that is less wordy. If n(n + 1)(n + 2)(n + 3) is a perfect square, it must be equal to ##(n + a)^4## for some integer a, and for all integers n.
Expanding the two expressions, we get ##n^4 + 6n^3 + 11n^2 + 6n## and ##n^4 + 4an^3 + 6a^2n^2 + 4a^3n + a^4##. Equate the coefficients to see that the original expression can't be a perfect square.
 
  • #6
139
2
Here's a different approach that is less wordy. If n(n + 1)(n + 2)(n + 3) is a perfect square, it must be equal to ##(n + a)^4## for some integer a, and for all integers n.
Expanding the two expressions, we get ##n^4 + 6n^3 + 11n^2 + 6n## and ##n^4 + 4an^3 + 6a^2n^2 + 4a^3n + a^4##. Equate the coefficients to see that the original expression can't be a perfect square.
Ah hello, thanks for the response

Can you tell me why you put it as (n+a)^4 ? surely if it's a perfect square it should be of form k^2 where k = n + a. we can't say if k will remain an integer if another root is taken to make it to the fourth power .
 
  • #7
33,271
4,975
Ah hello, thanks for the response

Can you tell me why you put it as (n+a)^4 ? surely if it's a perfect square it should be of form k^2 where k = n + a. we can't say if k will remain an integer if another root is taken to make it to the fourth power .
##(n + a)^2## won't give us the fourth powers that we have when we expand n(n + 1)(n + 2)(n + 3). ##(n + a)^4## is a perfect square (of ##(n + a)^2##).
 
  • #8
139
2
##(n + a)^2## won't give us the fourth powers that we have when we expand n(n + 1)(n + 2)(n + 3). ##(n + a)^4## is a perfect square (of ##(n + a)^2##).
I dunno, say for example that n(n+1)(n+2)(n+3) leads to a number like 9 ie 3*3 . this cannot be written as a fourth power in form (n+a)^4, can it ?
 
  • #9
33,271
4,975
I dunno, say for example that n(n+1)(n+2)(n+3) leads to a number like 9 ie 3*3 . this cannot be written as a fourth power in form (n+a)^4, can it ?
Can 9 be written as the product of four consecutive integers? If not, then it doesn't matter that 9 isn't a perfect fourth power.
 
  • #10
1,943
241
Here's a different approach that is less wordy. If n(n + 1)(n + 2)(n + 3) is a perfect square, it must be equal to ##(n + a)^4## for some integer a, and for all integers n.
Expanding the two expressions, we get ##n^4 + 6n^3 + 11n^2 + 6n## and ##n^4 + 4an^3 + 6a^2n^2 + 4a^3n + a^4##. Equate the coefficients to see that the original expression can't be a perfect square.
I really can't see why n(n + 1)(n + 2)(n + 3) = (n+a)^4 has to be true for all n. If it isn't true, than equating coefficients isn't valid.
Your method would also prove that n(n+1)(n+2)(n+5) can't be a perfect square, wich has an obvious counterexample.
 
Last edited:
  • #11
33,271
4,975
I really can't see why n(n + 1)(n + 2)(n + 3) = (n+a)^4 has to be true for all n. If it isn't true, than equating coefficients isn't valid.
Your method would also prove that n(n+1)(n+2)(n+5) can't be a perfect square, wich has an obvious counterexample.
This is part of a proof by contradiction. The original statement is that n(n + 1)(n + 2)(n + 3) can't be a perfect square. In the proof by contradiction, we assume that n(n + 1)(n + 2)(n + 3) is a perfect square, and look for a contradiction.
 
  • #12
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,338
5,135
Here's a different approach that is less wordy. If n(n + 1)(n + 2)(n + 3) is a perfect square, it must be equal to ##(n + a)^4## for some integer a
Shouldn't that be ##(n^2+a)^2##?
 
  • #13
Dick
Science Advisor
Homework Helper
26,258
618
I really can't see why n(n + 1)(n + 2)(n + 3) = (n+a)^4 has to be true for all n. If it isn't true, than equating coefficients isn't valid.
Your method would also prove that n(n+1)(n+2)(n+5) can't be a perfect square, wich has an obvious counterexample.
I agree completely. I don't think that is valid. One way to actually do this is to multiply ##n(n+1)(n+2)(n+3)## out and notice that it's pretty close to being the square of something. How close is it to being a square?
 
  • #14
33,271
4,975
Shouldn't that be ##(n^2+a)^2##?
Could be. I know that we have to have four factors of n, but I haven't worked the problem out.
 
  • #15
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,338
5,135
Could be. I know that we have to have four factors of n, but I haven't worked the problem out.
An arbitrary square is of the form ##(n^2 + a)^2##, but not necessarily ##(n+a)^4##.

And, as others have pointed out you are looking for a single solution for ##n, a##, not polynomial equality for all ##n##.
 

Related Threads on Prove that the product of 4 consecutive numbers cannot be a perfect square

Replies
4
Views
3K
Replies
14
Views
749
Replies
4
Views
1K
Replies
11
Views
1K
Replies
7
Views
12K
Replies
6
Views
1K
Replies
7
Views
7K
Replies
9
Views
9K
Replies
1
Views
7K
Top