P divides all the coefficients of the terms with degree <8

[mathmari]

Hey!

$f(x)=\sum_{i=0}^{7}{a_ix^i}, g(x)=\sum_{i=0}^{5}{b_ix^i}$, $a_i, b_i$ are integer coefficients.
The prime $p$ divides $a_0, a_1, a_2, a_3, a_4$ but it does not divide $a_5, a_6, a_7$. $p$ divides also $b_0, b_1, b_2$, but it does not divide $b_3, b_4, b_5$.
$h(x)=f(x)g(x)$
Show that at $h(x)$ $p$ does not divide the coefficient of $x^8$, but it does divide all the coefficients of the terms with degree $<8$.

We suppose that $p \mid c_8$.
$p \mid b_1 \Rightarrow p \mid a_7 b_1$
$p \mid b_2 \Rightarrow p \mid a_6 b_2$
$p \mid a_3 \Rightarrow p \mid a_3 b_5$
$p \mid a_4 \Rightarrow p \mid a_4 b_4$

$c_8=a_7b_1+a_6b_2+a_5b_3+a_4b_4+a_3b_5 \Rightarrow a_5 b_3=c_8-a_7b_1-a_6 b_2-a_4 b_4-a_3b_5$

$p \mid c_8-a_7b_1-a_6 b_2-a_4 b_4-a_3b_5 \Rightarrow p \mid a_5 b_3$
Since $p$ is a prime $p \mid a_5$ or $p \mid b_3$. Both are not true, so $p \nmid c_8$.

To show that $p$ divides all the coefficients of the terms with degree $<8$, what am I supposed to do? Do I have to calculate all the coefficients and find that $p$ divides them? Or is there an other way to show this?

 

[I like Serena]

Hey!

$f(x)=\sum_{i=0}^{7}{a_ix^i}, g(x)=\sum_{i=0}^{5}{b_ix^i}$, $a_i, b_i$ are integer coefficients.
The prime $p$ divides $a_0, a_1, a_2, a_3, a_4$ but it does not divide $a_5, a_6, a_7$. $p$ divides also $b_0, b_1, b_2$, but it does not divide $b_3, b_4, b_5$.
$h(x)=f(x)g(x)$
Show that at $h(x)$ $p$ does not divide the coefficient of $x^8$, but it does divide all the coefficients of the terms with degree $<8$.

We suppose that $p \mid c_8$.
$p \mid b_1 \Rightarrow p \mid a_7 b_1$
$p \mid b_2 \Rightarrow p \mid a_6 b_2$
$p \mid a_3 \Rightarrow p \mid a_3 b_5$
$p \mid a_4 \Rightarrow p \mid a_4 b_4$

$c_8=a_7b_1+a_6b_2+a_5b_3+a_4b_4+a_3b_5 \Rightarrow a_5 b_3=c_8-a_7b_1-a_6 b_2-a_4 b_4-a_3b_5$

$p \mid c_8-a_7b_1-a_6 b_2-a_4 b_4-a_3b_5 \Rightarrow p \mid a_5 b_3$
Since $p$ is a prime $p \mid a_5$ or $p \mid b_3$. Both are not true, so $p \nmid c_8$.

To show that $p$ divides all the coefficients of the terms with degree $<8$, what am I supposed to do? Do I have to calculate all the coefficients and find that $p$ divides them? Or is there an other way to show this?

Hii! ;)

I suggest to verify $c_7$ and make an argument for the others.

 

[mathmari]

Hii! ;)

I suggest to verify $c_7$ and make an argument for the others.

$c_7=a_7b_0+a_6b_1+a_5b_2+a_4b_3+a_3b_4+a_2b_5$

$p \mid b_0 \Rightarrow p \mid a_7 b_0$
$p \mid b_1 \Rightarrow p \mid a_6 b_1$
$p \mid b_2 \Rightarrow p \mid a_5 b_2$
$p \mid a_2 \Rightarrow p \mid a_2 b_5$
$p \mid a_3 \Rightarrow p \mid a_3 b_4$
$p \mid a_4 \Rightarrow p \mid a_4 b_3$

So $p \mid c_7$

What can I say for the others?

 

[Deveno]

Note that:

$\displaystyle c_k = \sum_{i = 0}^k a_ib_{k-i}$

We will take $b_j = 0$ for $j > 5$.

If $k \leq 4$, then every term is divisible by $p$ since $p$ divides all the "$a$'s".

This leaves just 3 coefficients to check: $c_5,c_6,c_7$. We could manually compute them, but that seems like...ugh...work.

But notice that as the "$a$'s" count up, the "$b$'s" count down. So if the b terms take care of the a terms that aren't covered, every term in the sum will be divisible by $p$. And the LOWEST $k$ for which this happens is when the i in:

$a_ib_{k-i}$ is greater than 4, and the k-i in said term is greater than 2, that is:

$k = i + (k-i) \geq 5 + 3 = 8$.

Since 5,6 and 7 are all less than 8, $c_5,c_6,c_7$ are all divisible by $p$.

 

[I like Serena]

$c_7=a_7b_0+a_6b_1+a_5b_2+a_4b_3+a_3b_4+a_2b_5$

$p \mid b_0 \Rightarrow p \mid a_7 b_0$
$p \mid b_1 \Rightarrow p \mid a_6 b_1$
$p \mid b_2 \Rightarrow p \mid a_5 b_2$
$p \mid a_2 \Rightarrow p \mid a_2 b_5$
$p \mid a_3 \Rightarrow p \mid a_3 b_4$
$p \mid a_4 \Rightarrow p \mid a_4 b_3$

So $p \mid c_7$

What can I say for the others?

As Deveno also remarked, we have low numbered $b_i$ that are divisible by $p$ or we have low numbered $a_i$ that are divisible by $p$.
For $c_7$ we have the worst case that still pans out.
So lower numbered $c_i$ will also be divisible by $p$.