# Problem Of The Week #401 Jan 15th, 2020

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#### anemone

##### MHB POTW Director
Staff member
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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#### anemone

##### MHB POTW Director
Staff member
Congratulations to Opalg for his partial correct solution!

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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