# Homework Help: Prove that the product of 4 consecutive numbers cannot be a perfect square

1. May 28, 2018

### Vriska

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Uniqueness of prime factors for a given number

3. 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.

2. May 28, 2018

### Vriska

Agh, I meant product !

3. May 28, 2018

### Staff: Mentor

I corrected the title.

4. May 28, 2018

### Vriska

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. May 28, 2018

### Staff: Mentor

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. May 28, 2018

### Vriska

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. May 28, 2018

### Staff: Mentor

$(n + a)^2$ 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. May 28, 2018

### Vriska

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. May 28, 2018

### Staff: Mentor

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. Jun 14, 2018

### willem2

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: Jun 14, 2018
11. Jun 15, 2018

### Staff: Mentor

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. Jun 15, 2018

### PeroK

Shouldn't that be $(n^2+a)^2$?

13. Jun 15, 2018

### Dick

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. Jun 15, 2018

### Staff: Mentor

Could be. I know that we have to have four factors of n, but I haven't worked the problem out.

15. Jun 15, 2018

### PeroK

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 $n$.