Let me start with two observations. 


* In the classification of quadratic forms with rational coefficients, one has the following statement: a quadratic form in five indeterminates represents $0$ over $\mathbb Q$ if and only if it represents $0$ over $\mathbb R$.
* For every $n\ge2$, there exists a division ring $R\_n$ of dimension $n^2$ over its center $\mathbb Q$. There is a 'norm' over $R\_n$, which is an $n$-form (homogeneous polynomial of degree $n$) with rational coefficients, multiplicative over $R\_n$. It is an example of $n$-form in $n^2$ indeterminates that represents $0$ over $\mathbb R$ but does not over $\mathbb Q$.



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> Question. Is it true that for every $n\ge2$, an $n$-form with rational coefficients in $n^2+1$ indeterminates represents $0$ over $\mathbb Q$ if and only if it represents $0$ over $\mathbb R$ ?
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A positive answer would say that the norm in a division ring is a maximal (in terms of the number of indeterminates) example of an $n$-form representing zero over $\mathbb R$ but not over $\mathbb Q$.


*Edit*. After Laurent's answer, I see that my question was close to Artin's conjecture, that Terjanian's example disproves. However, for fixed degree $n$, Artin's conjecture is true for almost every prime $p$: every $n$-form in $n^2+1$ indeterminates represents zero over $Q\_p$.



$n$-forms representing zero (versus division rings)
Take Terjanian's example of a form of degree 4 in 18 variables over $\mathbb{Q}$ without nontrivial zero over $\mathbb{Q}\_2$ (hence over $\mathbb{Q}$), as explained e.g. in Serre, Cours d'arithmétique, chap. 4. It is easy to see that this form represents zero over $\mathbb{R}$: in fact it is negative at the point (1,0,0,...) and positive at (1,1,1,0,0,...). Hence you get a 17-variable example by setting the last variable equal to zero.


