MHB Is (x^2+1) a maximal ideal in Z7[x]?

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The polynomial \( p(x) = x^2 + 1 \) in \( \mathbb{Z}_7[x] \) has no roots in \( \mathbb{Z}_7 \), indicating it is irreducible. Since \( p(x) \) is irreducible and has degree 2, the ideal generated by \( p(x) \), denoted as \( (p) \), is a prime ideal. In a principal ideal domain, all prime ideals are also maximal. Therefore, \( (p) \) is confirmed to be a maximal ideal in \( \mathbb{Z}_7[x] \.
Fernando Revilla
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I quote an unsolved question from Yahoo! Answers

Give a maximal ideal of Z7[x] with justification

I have given a link to the topic there so the OP can see my complete response.

P.S. I have posted there only a hint.
 
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Consider the polynomial $p(x)\in\mathbb{Z}_7[x]$, $p(x)=x^2+1$. This polynomial has no roots: $$p(0)=1,\;p(1)=2,\;p(2)=5,\;p(3)=3,\;p(4)=3,\;p(5)=5,\;p(6)=2$$ and has degree 2 which implies that $p(x)$ is irreducible, as a consequence $(p)$ is a prime ideal. But in a principal ideal domain all prime ideals are maximal, hence, $(p)$ is a maximal ideal.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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