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No idea how to proceed with this question:-
$P=2008^{2007}-2008$
$Q=2008^2+2009$
The remainder when $P$ is divided by $Q$ is

Any theorem or lemma used, just tell me what it is along with the solution

This question is from a regional maths contest.This question came and I couldn't make out any possible way to solve it. I thought of applying modulo but the values were far too large.

user452508
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  • Did you try anything yourself? It's unlikely that someone will give you a step by step solution given that you didn't give as any information as to what did you try,where did you get stuck,where this question is coming from and such. – kingW3 Jun 09 '17 at 13:27
  • Synthetic division won't work (as the divisor is a quadratic), but I'm sure you can find a generalisation of it by searching it up. What else have you tried? Where did you find the question? At what level is the question aimed at? I'm sure you can add extra context on your question to help people answering. – Toby Mak Jun 09 '17 at 13:28
  • In one sentence: show some context @sayan and put effort so people will ACTUALLY answer. – Toby Mak Jun 09 '17 at 13:31
  • Expanding the first few terms (x^2+x+1)^1003 using the binomial theorem will show a clear pattern, from there you can $x^{2007}-1$ from the expansion. – Toby Mak Jun 09 '17 at 13:38
  • By Maple the reminder is $4032066=2\times 3 \times 19 \times 113 \times 313$, maybe useful. – Amin235 Jun 09 '17 at 13:41

2 Answers2

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${\rm mod}\ a^{\large 2}\!+a+1\!:\,\ 0\equiv (a^{\large 2}\!+a+1\!)(a\!-\!1)\equiv a^{\large 3}-1\ $ so $\ \color{#C00}{a^{\large 3}\equiv 1}\ $ so $\ \color{#0a0}{a^{\large 3n}}\equiv (\color{#c00}{a^{\large 3}})^{\large n}\equiv \color{#c00}1^{\large n}\equiv \color{#0a0}1$

hence $\ \color{#0a0}{a^{\large 3n}}\!-a\,\equiv\, \color{#0a0}1\!-\!a.\,$ Yours is case $\,a = 2008,\ 3n= 2007\, (\equiv 2\!+0\!+\!0\!+\!7\equiv 0\pmod{\!\!3})$

Remark $ $ More generally, as above $\ a^{\large J}\! + a^{\large K}\! \equiv a^{\large J\,{\rm mod}\,\color{#c00} 3}+ a^{\large K\,{\rm mod}\, \color{#c00}3}\,\pmod{\!a^{\large 2}\!+a+1}$

For example $\ a^{\large 2}\!+\!a\!+\!1\mid a^{\large J}\! +\! a^{\large K}\! +\! a^{\large L}\ $ if $ \ \{J,K,L\}\equiv \{2,1,0\}\pmod{\!3}$

Seee this answer for further related examples.

Bill Dubuque
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If you put $x=2008$ you have to divide $$\frac{x^{2007}-x}{x^2+x+1}$$ Now to make calculations easier lets divide by $x^3-1=(x-1)(x^2+x+1)$ which is easy since $x^3-1\mid x^{2007}-1$ $$\frac{x^{2007}-x}{x^3-1}=x^{2004}+x^{2001}+x^{1998}+\cdots+x^3+1+\frac{1-x}{x^3-1}$$ Multiplying by $x-1$ you get that $$\frac{x^{2007}-x}{x^2+x+1}=(x-1)(x^{2004}+\cdots+x^3+1)+\frac{1-x}{x^2+x+1}$$ Now since having a negative number as remainder is not convenient use the fact that $1-x\equiv Q-(1-x)\pmod{Q}$

Now just replace $Q$ with the actual value.

kingW3
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  • +1 if you put a link to the Remainder Theorem or give a little explanation of the Remainder Theorem. Without that, it is near impossible to find more information on the topic. – AlgorithmsX Jun 09 '17 at 14:40