# Product of 9 consecutive numbers cannot be partitioned into two sets with equal product. Where is my flaw in the proof?

I have the following problem:

Show that 9 consecutive integers cannot be partitioned into two sets such that the products of the first and second set are equal.

I know this question has been asked multiple times. Nearly all of the answers argue with prime factorization and I was just wondering where the flaw in my argumentation is:

Suppose such a partition exists. Then the product of the 9 consecutive integers needs to be a perfect square $$n^2$$. The product can also be represented as $$k(k+1)\dots{}(k+8)$$ which when multiplied out yields a polynomial of the form $$k^9 + \text{polynomial of degree} \leq 8.$$ In order for this to be a perfect square, we would need to be able to represent the polynomial in the form $$(\dots{})^2$$. Since the degree of the polynomial is odd, we cannot do this and hence such a partition does not exist.

Edit: I know that there is a theorem (Erdos-Selfridge) that states the product of consecutive integers can never be a perfect power $$x^l$$. However, I was wondering if my argument about the even/odd property for this special case holds.

• A non-square polynomial can still have square numbers as values.
– Karl
Commented Jan 8, 2023 at 13:08
• @Karl Ok, I see. My argument only states that it cannot be square for all $k \in \mathbb{N}$ (because for this we would need the factorization which is impossible) but it could be the case that there exists a$k$ for which the polynomial would be square. Thus the problem lies in wrong use of $\forall, \exists$. Commented Jan 8, 2023 at 13:20
• That sounds right, but how do you know the polynomial must factor as the square of a polynomial in order to always take square-number values? That's not obvious to me.
– Karl
Commented Jan 8, 2023 at 19:43
• @user376343 Are you claiming $8^9$ is a perfect square? Because it is not. Commented Jan 8, 2023 at 23:11
• @FedericoPoloni right, I made a typo, thanks for pointing out. Commented Jan 8, 2023 at 23:29

## 2 Answers

Consider two distinct questions about polynomials:

1. Can a polynomial of odd degree be a square of another polynomial?
2. Can a polynomial of odd degree take a value which is a square?

Your attempted proof depends on a negative answer to (2), but in fact (2) is true, for example $$k^9+k^4+1$$ is of odd degree, but when $$k=2$$ takes the value $$529=23^2$$.

Instead of proving, more specific to the initial question why your approach does not hold. You write $$k(k+1)…(k+8)$$ and then state that $$k^9+\text{polynomial of degree}≤8$$ cannot be a square because the degree is odd. But, you might also write $$k(k+1)…(k+7)(k+i)$$ which is a polynomial with the same odd degree but you will certainly be able to find an $$i$$ for which the product is a square (i.e. $$k(k+1)…(k+7)=(k+i)$$). Not sequentially integers, but the claim was about the degree of the polynomial.

Even for a polynomial with odd degree one: $$k+1$$ will have many solutions so $$k+1$$ is a square.