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

Question

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.

Answer 1

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 $529=23^2$.

Answer 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.