MHB What is the Smallest Possible Value of a in This Polynomial Pattern?

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The problem involves finding the smallest positive value of \( a \) for a polynomial \( P(x) \) with integer coefficients that satisfies specific conditions at the integers 1, 3, 5, and 7, equating to \( a \), while at 2, 4, 6, and 8, it equates to \(-a\). The polynomial must exhibit a pattern where it alternates between these values at specified points. A suggested solution indicates that the smallest possible value of \( a \) is 4, derived from analyzing the polynomial's behavior and constraints. The discussion highlights the importance of polynomial properties and integer coefficients in determining the solution. Overall, the problem emphasizes the relationship between polynomial roots and their evaluated outputs.
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Here is this week's POTW:

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Let $a>0$ and $P(x)$ be a polynomial with integer coefficients such that

$P(1)=P(3)=P(5)=P(7)=a$ and

$P(2)=P(4)=P(6)=P(8)=-a$.

What is the smallest possible value of $a$?

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Congratulations to Opalg for his partial correct solution!(Cool)

You can find the suggested solution as shown below:
Because 1, 3, 5 and 7 are roots of the polynomial $P(x)-a$, it follows that

$P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$,

where $Q(x)$ is a polynomial with integer coefficients. The identity must also hold for $x=2,\,4,\,6$ and $8$, thus,

$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8)$

Therefore $315=\text{lcm} (15,\,9,\,105)$ divides $a$, that is $a$ is an integer multiple of 315.

Let $a=315A$. Because $Q(2)=Q(6)=42A$, it follows that $Q(x)-42A=(x-2)(x-6)R(x)$, where $R(x)$ is a polynomial with integer coefficients.

Because $Q(4)=-70A$ and $Q(8)=-6A$, it follows that $-112A=-4R(4)$ and $-48A=12R(8)$, that is $R(4)=28A$ and $R(8)=-4A$.

Thus, $R(x)=28A+(x-4)(-6A+(x-8)T(x))$, where $T(x)$ is a polynomial with integer coefficients.

Moreover, for any polynomial $T(x)$ and any integer $A$, the polynomial $P(x)$ constructed this way satisfies the required conditions. The required minimum is obtained when $A=1$ and so $a=315$.
 
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