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This is a problem I have faced great difficulty with. (teacher's challenge question)

Suppose $G(n)$ is a monic polynomial with integer coefficients in which $G(0)=31$. Also, suppose that the distinct integers $b_1,...,b_q$ satisfy $G(b_1)=...=G(b_q)=65$

1) Find the maximum possible value of $q$ (over all $G$)

I got that $q=4$ since the maximum number of terms that can multiplied together to give $-34$($31-65$) is 4.

2) Determine all G for which this maximum is achieved ($q=4$)

I am not quite sure how to approach this problem, please help I am really frustrated and this is really important for me to be able to solve! (I have a test and I should be able to do questions similar to these...)

P.S. I didn't put up my work for question 1 because that's not what I am really concerned about. My central question is number 2.

Hiro
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  • I can't personally think of a good title for this question right now, but its worth mentioning you very likely don't want to include "algebraic geometry" in the title or the tags. Algebraic Geometry is a very specific area of math that this isn't really related to. – Mark Schultz-Wu Nov 10 '16 at 01:01
  • Thanks for the comment, I changed it to suit the topic. – Hiro Nov 10 '16 at 01:02
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    How did you get $q=4$? Can you write down one possibility for $G$ using those 4 values? – ChocolateAndCheese Nov 10 '16 at 01:44
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    I found $q=4$ by doing this: $(n-b_1)(n-b_2)...(n-b_q)=G(n)-65$ Then plugging $n$ for $0$ and I get $(-b_1)(-b_2)...(-b_q)=31-65$ =>$(-b_1)(-b_2)...(-b_q)=-34$ => What is the maximum number of distinct integers that could be multiplied to give $-34$? An example, $-34=1-1217$ The maximum is 4, which means that $q=4$ – Hiro Nov 10 '16 at 01:54
  • think $-34=1-1217=1-1-2-17$. So one polynomial is $H(n)=(n-1)(n+1)(n-2)(n-17)+65$, and another polynomial is $K(n)=(n-1)(n+1)(n+2)(n+17)+65$. Are there any others? Note that $H(n)=n^4-19n^3+33n^2+19n+31$, and $K(n)=n^4+19n^3+33n^2-19n+31$. – Mirko Nov 10 '16 at 02:37
  • @Mirko Would you have to substitute $H(n)$ for $G(n) -65$ and $K(n)$ for $G(n)-65$ in order to write the equation in terms of $G(n)$? Or did you already do so by adding the $65$ to both functions? – Hiro Nov 10 '16 at 02:44
  • I just used what you wrote, $(n-b_1)(n-b_2)...(n-b_q)=G(n)-65$. Adding $65$ to both sides gives $(n-b_1)(n-b_2)...(n-b_q)+65=G(n)$. Taking $(b_1,b_2,b_3,b_4)=(1,-1,2,17)$ produces $G(n)=(n-1)(n+1)(n-2)(n-17)+65$. On the other hand, taking $(b_1,b_2,b_3,b_4)=(1,-1,-2,-17)$ produces $G(n)=(n-1)(n+1)(n+2)(n+17)+65$. Since you get two different versions of $G(n)$, I just preferred to call one of them $H(n)$ and the other $K(n)$. – Mirko Nov 10 '16 at 02:53
  • So is my way of approaching this problem correct, since I see that you're using what I wrote to solve the next question~ If yes, I'll be relieved lol/ if not, could you explain why not? – Hiro Nov 10 '16 at 02:56
  • your way is correct ... you just need to explain it to yourself one more time, to understand it better. – Mirko Nov 10 '16 at 03:04
  • Is this not the same as http://math.stackexchange.com/questions/2001524/polynomials-in-trouble posted recently? – Gerry Myerson Nov 10 '16 at 03:41
  • @GerryMyerson I wasn't aware there's a similar question to this, I guess I'll just delete this one then, but thanks for the help guys! – Hiro Nov 10 '16 at 03:51

1 Answers1

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If $P(n)=G(n)-65$ then $P(b_1)=...=P(b_q)=65-65=0$, and
$P$ is monic just like $G$ is.
($G(n)$ cannot be a constant polynomial since $G(0)=31\not=65$.
Hence $\deg(G)=\deg(P)\ge1$.)

Hence $P(n)=(n-b_1)^{a_1}(n-b_2)^{a_2}...(n-b_q)^{a_q}$ and
$G(n)=P(n)+65=(n-b_1)^{a_1}(n-b_2)^{a_2}...(n-b_q)^{a_q}+65$
for some positive integers $a_1,a_2,...,a_q$.

Then $31=G(0)=(-b_1)^{a_1}(-b_2)^{a_2}...(-b_q)^{a_q}+65$,
hence $(-b_1)^{a_1}(-b_2)^{a_2}...(-b_q)^{a_q}=31-65=-34$.

Now you would like to factor $-34$ into distinct integers,
as many as possible.
One option is $-34=1\cdot(-1)\cdot2\cdot17$, and
another is $-34=1\cdot(-1)\cdot(-2)\cdot(-17)$.

This gives two possibilities for $G(n)$. One of them is when
$(-b_1)(-b_2)(-b_3)(-b_4)=-34=1\cdot(-1)\cdot2\cdot17$, then
$G(n)=(n-b_1)(n-b_2)(n-b_3)(n-b_4)+65=$
$=(n-1)(n+1)(n-2)(n-17)+65=$
$=n^4-19n^3+33n^2+19n+31$. Here $q=4$.

The other possibility is when
$(-b_1)(-b_2)(-b_3)(-b_4)=-34=1\cdot(-1)\cdot(-2)\cdot(-17)$, then
$G(n)=(n-b_1)(n-b_2)(n-b_3)(n-b_4)+65=$
$=(n-1)(n+1)(n+2)(n+17)+65=$
$=n^4+19n^3+33n^2-19n+31$. Here again $q=4$.

Mirko
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