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

In summary: 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. Then, by contradiction, we get that n(n+1)(n+2)(n+3) = (n+a)^4.
  • #1
Vriska
138
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 don't know 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.
 
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  • #2
Agh, I meant product !
 
  • #3
Vriska said:
Agh, I meant product !
I corrected the title.
 
  • #4
DrClaude said:
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
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.
 
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  • #6
Mark44 said:
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
Vriska said:
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
Mark44 said:
##(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
Vriska said:
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
Mark44 said:
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, which has an obvious counterexample.
 
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  • #11
willem2 said:
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, which 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.
 
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  • #12
Mark44 said:
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
willem2 said:
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, which 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?
 
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  • #14
PeroK said:
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
Mark44 said:
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##.
 

What is meant by "product of 4 consecutive numbers"?

The product of 4 consecutive numbers refers to the result of multiplying 4 numbers that come one after the other in a sequence. For example, the product of 2, 3, 4, and 5 is 2*3*4*5=120.

What is a perfect square?

A perfect square is a number that is the product of two equal integers. For example, 9 is a perfect square because it is equal to 3*3.

Why can't the product of 4 consecutive numbers be a perfect square?

This is because the product of 4 consecutive numbers will always have an odd number of factors. A perfect square, on the other hand, will have an even number of factors. Since the number of factors is different, the product of 4 consecutive numbers cannot be a perfect square.

Can you provide an example to prove this statement?

Yes, let's take the numbers 2, 3, 4, and 5 as an example. Their product is 2*3*4*5=120. This number has the factors 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120. As you can see, there are 16 factors, which is an even number. Therefore, 120 cannot be a perfect square.

Is there a general proof for this statement?

Yes, there is a general proof for this statement. It involves using the concept of prime factorization and showing that the product of 4 consecutive numbers will always have an odd number of prime factors, while a perfect square will have an even number of prime factors. This means that the two numbers can never be equal, and therefore the product of 4 consecutive numbers cannot be a perfect square.

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