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

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!

• Well I suppose the question to ask is what have you learned i.e., what level was this course at?
– Mike
May 20 at 0:31
• 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.
– Mike
May 20 at 0:46
• Thanks... I need to find the polynomial.... May 20 at 2:20
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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.

• Thank you!. Is there a different way??? I don't know about Galois Group... May 20 at 2:35
• This does need to be elaborated on, please, it really does not answer the question as is...
– Mike
May 20 at 17:28
• 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. May 20 at 17:42
• As the OP themself mentioned, they didn't study Galois groups....I think an elementary solution is what is called for here
– Mike
May 20 at 17:44
• ...or a more elementary solution, if you would, Sir....
– Mike
May 20 at 18:35