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I'm preparing for an entrance exam and got stuck on a question.

Let $f(x)$ be a polynomial of degree greater than $1$. If $f(x)$ is divided by $x-a$, then $f(a)$ is the remainder.

Q1) Let $f(x) = x^{2013} +1$, then remainder when $f(x)$ is divided by $x^2 -1$ is:

A) $x-1$

B) $x+1$

C) $0$

D) $1$

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  • $\begingroup$ Please show your working so far :) $\endgroup$ – Shaun Dec 11 '13 at 11:02
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Hint: Write $f(x)= q(x)(x^2-1)+ax+b$ and evaluate at $x=\pm 1$ to find $a$ and $b$.

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  • $\begingroup$ This is a general technique for finding the remainder in the division by a polynomial whose roots are known. $\endgroup$ – lhf Dec 11 '13 at 11:49
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You could, in this instance, just replace every occurrence of $x^2$ by $1$. Since $2013$ is odd, $x^{2013}=(x^2)^{1006}\cdot x$ just leaves $x$.

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$x^{2013}+1=x\left(x^{2012}-1\right)+x+1=x\left(x^{2}-1\right)\left(x^{2010}+x^{2008}+\cdots+x^{2}+1\right)+x+1$

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$$T_{n}=x^{n}+1\\ T_{2013} = x^{2013}+1\equiv x^{2013}+1-x^{2011}(x^2-1) = x^{2011}+1\pmod {x^2-1}\\ T_n\equiv T_{n-2}\equiv T_{n\space\bmod\space2}\pmod {x^2-1}\\ T_{2013}\equiv T_1=x+1\pmod {x^2-1}$$

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  • $\begingroup$ Use \equiv to write the $\equiv$ symbol. It will make your answer more understandable. $\endgroup$ – Peter Phipps Dec 11 '13 at 20:47

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