I am struggling with the following question:
Let $\mathbb{F}_9$ be the field with nine elements. How many triples $a,b,c$ of elements in $\mathbb{F}_9$ are there, such that $a^2 + b^2 = c^2$ ?
I suppose the brute force way is to go through all combinations $a,b,c$ of elements in $\mathbb{F}_9$, and check algebraically which ones satisfy the given equation $a^2 + b^2 = c^2$. However, with $9$ lements in the field, this gives $9^3$ possible triples, which is way too many to check.
How can I solve this without using brute force ?
Another idea I had is to identity $\mathbb{F}_9$ with the quotient $\mathbb{Z}_3[x]/(x^2 + x + 2)$, where $(x^2 + x + 2)$ is the ideal generated by the polynomial $x^2 + x + 2$, which is irreducible in $\mathbb{Z}_3[x]$. However, this still gives $9^3$ polynomials corresponding to triples $a,b,c$ in $\mathbb{F}_9$, again giving too many possibilities to check by brute force whether the equation $a^2 + b^2 = c^2$ is satisfied.
Thanks!
I found brute force was not as bad as anticipated.
Say $\mathbb F_9=\mathbb Z_3[i]$, where $i^2=2$.
The elements of $\mathbb F_9$ are then $0, 1, 2, i, 1+i, 2+i, 2i, 1+2i, $ and $2+2i$.
Their squares are $0, 1, 2, i,$ and $2i$. The non-squares are $1+i, 2+i, 1+2i$, and $2+2i$.
So there are five choices for $a^2$ and five choices for $b^2$.
The following seventeen $(a^2,b^2)$ pairs satisfy $a^2+b^2=c^2$ for some $c\in\mathbb F_9$:
$(0,0), (0,1), (0,2), (0,i), (0,2i), (1,0), (2,0), $
$(i,0), (2i, 0), (1,1), (1, 2), (2,1), (2,2), (i,i), (i, 2i), (2i, i), $ and $(2i, 2i)$.
The following eight $(a^2,b^2)$ pairs do not satisfy $a^2+b^2= c^2$ for some $c\in\mathbb F_9$:
$(1,i), (1,2i), (i,1), (2i,1), (2,i), (i,2), (2,2i), (2i,2)$.