What is the remainder of dividing the $P(x)$ polynomial by $x^2+1$?

Question

If $P(x)= (x+1)^{2015}+(x-1)^{2015}$ , what is the remainder of dividing the $P(x)$ polynomial by $x^2+1$?

A) $-2^{2015}x$

B) $-2^{1008}x$

C) $2^{1008}x$

D) $2^{2008}$

E) $1$

My attepmts:

Problem is equivalent to $$P(x) \mod (x^2+1)=?$$

Let $x=1$, we have $P(1) \mod (1^2+1)=0$ which imply $R(1)=0$. But no answer matches this. Where is the problem? I can not see my mistake.

Answer 1

Drop the modular arithmetic for a second, and take one step back, so that it's easier to see what's going on when you substitute for $x$. We have that $$ P(x)=Q(x)(x^2+1)+ax+b $$ for some polynomial $Q$ and real coefficients $a,b$. We are after $ax+b$.

Note that inserting $x=1$ into this doesn't get us closer, as that gives us $$ 2^{2015}=2Q(2)+a+b $$ and we still don't have a handle on $Q$. The only thing this tells us is that $a+b$ is even.

On the other hand, inserting $x=i$ or $x=-i$ will make $x^2+1=0$, and the whole $Q$ term vanishes, leaving you with two equations you can solve for $a$ and $b$.

Answer 2

$P(x)= 2(x^{2015}+ \binom{2015}{2} x^{2013}+\binom{2015}{4}x^{2011}+ \cdots \binom{2015}{2014}x)$

Using the remainder theorem, we have $P(i)$ as the remainder when $P(x)$ is divided by $x^2+1$.

Complete it from here. You may have to use some identity of binomial coefficients.

Answer 3

Hint:

$$(x-1)^2\equiv-2x\pmod{x^2+1}$$

$$(x-1)^3\equiv-2x(x-1)=2x-2x^2\equiv2x-(-2)$$

$$\implies(x-1)^4\equiv(-2x)^2\equiv-4$$

$$(x-1)^{4n+3}=((x-1)^4)^n(x-1)^3\equiv(-4)^n2(x+1)\equiv(-1)^n2^{2n+1}(x+1)$$

Similarly for $(x+1)^{4n+3}$

Here $4n+3=2015\iff n=?$

Answer 4

Let $$P(x)=(x^2+1)Q(x)+Ax+B~~~(*)$$ Put $x=i$ in above to get $$(i+1)^{2015}+(i-1)^{2015}=Ai+B \implies (1+i)^{2015}-(1-i)^{2015}=Ai+B ~~~(1)$$ By putting $x=-i$ we get $$(-i+1)^{2015}+(-i-1)^{2015}=-Ai+B \implies -[(1+i)^{2015}-(1-i)^{2015}])~~~(2)$$ Adding (1) and (2) get $B=0$. Then from (1) $$2^{2015/2} 2i \sin(2015 \pi/4)= A i \implies A=2^{2015/2}~~ 2 \frac{-1}{\sqrt{2}}=2^{(2015-1)/2}\times -2=-2^{1008}$$ So the reequired remainder from $(*)$ is $$-2^{1008}x$$. The option (B) is correct.