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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.

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    $\begingroup$ A non-square polynomial can still have square numbers as values. $\endgroup$
    – Karl
    Commented Jan 8, 2023 at 13:08
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    $\begingroup$ @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 $\forall, \exists$. $\endgroup$
    – LeoMinor
    Commented Jan 8, 2023 at 13:20
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    $\begingroup$ 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. $\endgroup$
    – Karl
    Commented Jan 8, 2023 at 19:43
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    $\begingroup$ @user376343 Are you claiming $8^9$ is a perfect square? Because it is not. $\endgroup$ Commented Jan 8, 2023 at 23:11
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    $\begingroup$ @FedericoPoloni right, I made a typo, thanks for pointing out. $\endgroup$
    – user376343
    Commented Jan 8, 2023 at 23:29

2 Answers 2

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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 $529=23^2$.

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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.

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