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I am trying to solve this question to study for my Number Theory final exam

QUESTION: Find a polynomial of the form $F(x,y,z)$ of degree 3 such that $F(a,b,c) \equiv 0 \pmod{5}$ iff $a,b,c \equiv 0\pmod{5}$

My attempt: "$\longrightarrow$" Let $f=x + y+ z$, Then take $f(a,b,c)= a + b + c \equiv 0\pmod{5} $ .

Then we know that $5| a + b + c$, thus the polynomial must be re-written as $5a + 5b + 5c$. Hence, when taken modulo 5, $a,b,c \equiv 0\pmod{5}$

PD: I Don't seem to understand how to prove the converse, I know Hensel's lemma, but not too sure how to apply it I still need to find the degree 3 polynomial.... Please help, I am a beginner, and I'm still learning.

UPDATE PLEASE READ: I had an Idea, please tell me what you guys think.

Let $f(x,y,z)=x^3 + y^3 + z^3$

So, suppose that $a,b,c \equiv 0$ mod 5. Then we can rewrite $a,b,c$ as $5k,5t,5m$. Then, when we plug these in the original equation, we will get: $f(5k,5t,5m)=125k + 125t + 125m$ which is $\equiv 0 $ mod 5.

Then, Supp. that $f(a,b,c) \equiv 0 $ mod 5. Then we can write $a^3 = 5t+ (-b^3 - c^3)$. But, if we rewrite b , and c in the same way, we will get that all contain a $5*$(some variable). Thus, in $a^3 = 5t + (-b^3 -c^3)$ we are able to factor out a 5, therefore we will get that $a^3 = 5*(something)$ which implies $a \equiv 0$ mod 5 . Do the same for b and c....

What do you guys think? Anything helps. Thank you in advance!

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    $\begingroup$ Well I suppose the question to ask is what have you learned i.e., what level was this course at? $\endgroup$
    – Mike
    May 20 at 0:31
  • $\begingroup$ For just two variables $a$ and $b$, and degree-$2$ it's easier: $F(a,b)=a^2+2b^2$ would do in that case. $\endgroup$
    – Mike
    May 20 at 0:46
  • $\begingroup$ Thanks... I need to find the polynomial.... $\endgroup$ May 20 at 2:20
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    – Glorfindel
    May 20 at 16:19

1 Answer 1

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I suppose that this is just a hint. First, the natural environment for this is $\Bbb F_5=\Bbb Z/5\Bbb Z$. You’re looking for a cubic form in three variables with no non-trivial zeros in $\Bbb F_5$.

Consider the (unique) cubic extension $k$ of your base field, find an explicit description of it for your computations. Now let $\{1=b_1,b_2,b_3\}$ be an $\Bbb F_5$-basis of $k$, and look at a general element $g=A+Bb_2+Cb_3$. Then, describe the (field-theoretic) Norm $\mathbf N^k_{\Bbb F_5}(g)$ of $g$, which you can describe as the product of $g$ with its two conjugates (*), or as the constant term of the $\Bbb F_5$-minimal polynomial of $g$. You’ll get a cubic form in the variables $A,B,C$. Now it’s a fact about the Norm that it’s nonzero for nonzero arguments, that is, elements of $k^\times$. And there you are.

(*) You treat $A,B,C$ here as elements of the base field $\Bbb F_5$, so not affected by elements of the Galois group.

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    $\begingroup$ Thank you!. Is there a different way??? I don't know about Galois Group... $\endgroup$ May 20 at 2:35
  • $\begingroup$ This does need to be elaborated on, please, it really does not answer the question as is... $\endgroup$
    – Mike
    May 20 at 17:28
  • $\begingroup$ Do you want me to do the answer out, @Mike ? I can, and it won’t take very long, but it will deprive the OP of all fun merely to read what I’ve written. $\endgroup$
    – Lubin
    May 20 at 17:42
  • $\begingroup$ As the OP themself mentioned, they didn't study Galois groups....I think an elementary solution is what is called for here $\endgroup$
    – Mike
    May 20 at 17:44
  • $\begingroup$ ...or a more elementary solution, if you would, Sir.... $\endgroup$
    – Mike
    May 20 at 18:35

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