# Let $f(x) =7x^{32}+5x^{22}+3x^{12}+x^2$. Then find its remainder in the following cases.

Let $f(x) =7x^{32}+5x^{22}+3x^{12}+x^2$.

(i) Then find the remainder when $f(x)$ is divided by $[x^2+1]$.

(ii) Also find the remainder when $xf(x)$ is divided by $[x^2+1]$.

Given both the remainders will be of the form $4(ax+b)$.

The 'polynomial long division method' can be applied, but that won't be logical enough, because its too long a process.

Can you tell me any other shorter & easier process?

Thanks.

• Please use proper punctuation, space after period, and no space before. – Gregory Grant May 12 '15 at 21:31
• What does "not logical enough" mean? – SalmonKiller May 12 '15 at 21:31
• @salmonkiller it means that OP will have to do 16 'steps' of long division, which is long for a problem. – Steven Stadnicki May 12 '15 at 21:33
• A broad hint: what is $x^2$ equivalent to $\pmod {x^2+1}$? – Steven Stadnicki May 12 '15 at 21:34
• @salmonkiller If i use long division it would be too long a process,i think.So i'm searching for a better process & want help. – Number May 12 '15 at 21:34

For the first case, note that only even powers of $x$ are involved, so you can substitute $y=x^2$. In this case you should find the remainder of $7y^{16}+5y^{11} + 3y^6 + y$ divided by $y+1$, i.e. $$7(-1)^{16}+5(-1)^{11} + 3(-1)^6 + (-1) = 4$$

So you have $f(x) = q(x)(x^2+1)+4$, where $q(x)$ is some polynomial.

From this follows that $xf(x) = xq(x)(x^2+1) +4x$, so the second remainder is $4x$.

• Good answer! In this case one definitely doesn't need to appeal to the machinery I did in my answer :). – Peter Woolfitt May 12 '15 at 21:49

(i) Note that $f(i)=7-5+3-1=4$, so the remainder is $4$.

(ii) Note that $if(i)=7i-5i+3i-i=4i$, so the remainder is $4x$.

This method uses the isomorphism $\dfrac{\mathbb{Z}[x]}{(x^2+1)}\simeq\mathbb{Z}[i]$, with $\varphi(x)=i$.

• Can you please expand and maybe mention the method? – SalmonKiller May 12 '15 at 21:34
• Why did you choose $x=i$ and not any other (real or complex) value? – barak manos May 12 '15 at 21:37
• @barakmanos I think that Peter Woolfitt used the isomorphism $R[x]/(x^2+1) = R[i] = C$. This is a little abstract approach, for people who is not comfortable with complex numbers. – Crostul May 12 '15 at 21:39
• @Crostul: That does not answer my question. – barak manos May 12 '15 at 21:41
• @barakmanos I did as Crostul suggested, and have updated my post. – Peter Woolfitt May 12 '15 at 21:44

Hint $\ {\rm mod}\,\ x^2\!+1\!:\,\ \color{#c00}{x^2\equiv -1}\,\Rightarrow\, f(\color{#c00}{x^2})\equiv f(\color{#c00}{-1}),\$ $\,xf(\color{#c00}{x^2})\equiv xf(\color{#c00}{-1}),\$ for $\ f\in\Bbb Z[x]$