# Find $a\in \mathbb{Z}_5[x] /I$ so $a^3=x^2+1+I$

Let $$f(x)=x^3+3x^2+3\in\mathbb{Z}_5[x]$$ and $$I=\langle f(x)\rangle$$ and $$R=\mathbb{Z}_5[x] /I$$.

I got stuck in the last exercise:

Find $$a\in R$$ so $$a^3=x^2+1+I$$.

What is the right approach of finding such $$a$$ without guessing?

• Did you mean $R = \mathbb{Z}_5[x]/I$? – Chessanator Jul 12 '19 at 20:25

Note that in $$\mathbb{Z}_5[x]/I$$, because of the way the polynomial $$f(x)$$ was chosen, we have that $$(x+I)^3= -3x^2 - 3 +I = -3 (x^2 +1 +I)$$. So if we were allowed to write $$a = \frac{x+I}{\sqrt[3]{-3}}$$ we'd have our answer.
Can you find a cube root of $$-3$$ in $$\mathbb{Z}_5$$? You can do a number of manipulations working mod $$5$$ to make this easier.
• how did you know to start using $(x+I)$? – vesii Jul 12 '19 at 20:45
• I saw the $3(x^2+1)$ in the polynomial $f$ as something to aim for. – Chessanator Jul 12 '19 at 20:48