At 1st we consider some weak statement of Chevalley–Warning theorem for any finite field: If $f$ is a homogeneous polynomial of degree $d$ with $n$ independent variables over a finite field $F$. Then if $ n > d $ then there is a non trivial solution of this homogeneous polynomial in $ F^{n} / \{0,0,...,0\} $.


Now if we consider $f$ is a homogeneous polynomial of degree $d$ with $n$ independent variables over the rational field $\mathbb{Q}$ and if $ n > d $ does there exist always a nontrivial solution in $ \mathbb{Q}^{n} / \{0,0,...,0\} $? The answer is no when $r$ is even as for an example $ x\_{1}^{d} + x\_{2}^{d}+.....+ x\_{n}^{d} = 0 $ have only one solution $\{0,0,...,0\} $ in $ \mathbb{Q}^{n}$.


My question is when $ d $ is odd does there exist always a nontrivial solution in $ \mathbb{Q}^{n} / \{0,0,...,0\} $? Describe when $ d = 3 $ and $ n= 5 $ especially.



Chevalley–Warning theorem for rational field $\mathbb{Q} $
The condition that $d$ is odd just implies that there is no real obstruction to the existence of rational points, but there could still be $p$-adic obstructions for some prime $p$.


As an example, let $p$ be a prime and $a$ an integer which is coprime to $p$ for which $a \bmod p$ is not a cube (this necessarily implies that $p \equiv 1 \bmod 3$).


Consider the cubic surface:
$$X: \quad x\_1^3 - a x\_2^3 + p(x\_3^3 - a x\_4^3) = 0.$$


It is a fun exercise to show that this has no non-trivial rational solution; the proof actually shows that there is no non-trivial $p$-adic solution. Note that there is obviously a non-trivial solution mod $p$, e.g. $(0,0,1,1) \bmod p$, as predicted by Chevalley-Warning.


A more interesting question is whether having a real point and a $p$-adic point is sufficient to guarantee a rational point; again this is not true in general, with smooth cubic surfaces giving counter-examples (this is called the Hasse principle in the literature). A famous example, due to Cassels and Guy, is:
$$5x\_1^3 + 12x\_2^3 + 9x\_3^3 + 10x\_4^3 = 0.$$
Proving that there is no non-trivial rational point in this case is not so elementary (it requires cubic reciprocity). From a modern perspective there is a Brauer--Manin obstruction to the Hasse principle in this case.


I suspect from the question you may have a specific example in mind you are studying. If you allow many variables (roughly $d2^d$) then the circle method can actually show that the Hasse principle holds, which may be sufficient for your application.


**Addendum:** This is a summary of the alternative construction given in the comments, written up for completeness.


Let $d \in \mathbb{N}$. Let $D$ be a division algebra over $\mathbb{Q}$ of dimension $d^2$. This can be shown to exist using the fundamental exact sequence for the Brauer group of $\mathbb{Q}$ from class field theory, combined with the fact that the period = index for central simple algebras over number fields.


Consider the [reduced norm](https://en.wikipedia.org/wiki/Central_simple_algebra) on $D$. After choosing a $\mathbb{Q}$-basis for $D$, this defines a homogeneous polynomial $N$ of degree $d$ in $d^2$ variables. We claim that the equation $N(\mathbf{x}) = 0$ has no non-trivial solution.


To see this, we use the fact that an element of a central simple algebra has reduced norm $0$ if and only if it is not invertible. But by construction $D$ is a division algebra, hence every non-zero element is invertible, hence only $\mathbf{0}$ has $0$ norm, as claimed.


Note: Division algebras satisfy the Hasse principle, hence this failure of existence of rational solutions is explained by a failure of existence of $p$-adic solutions for some $p$.


It is possible, though painful, to make these norm form equations explicit. The first issue is writing down explicit division algebras, which Will Sawin did in the comments.


As an example, consider the quaternion algebra $(a,b)$. The reduced norm here is just the [Pfister form](https://en.wikipedia.org/wiki/Pfister_form):
$$x\_1^2 - ax\_2^2 - bx\_3^2 + abx\_4^2.$$
Taking $b = p $ to be an odd prime and $a<0$ such that $a \bmod p$ is a quadratic non-residue, one obtains an example which does not non-trivially represent zero $p$-adically, but has real solutions. (The above cubic surface example is a variant of this equation).


