= Solution
The multiplicative <norm> first rules out <zero divisors>: if $xy=0$ and $x\neq0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.
It is worth checking that the printed description of a ring with an underlying real <vector space> really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, and
$$
|L_x(y)-L_x(z)|=|x(y-z)|=|x||y-z|.
$$
It is therefore <continuous>. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=tL_x(y)$ for every real $t$. The same argument holds in the other variable. This is <automatic real bilinearity from a multiplicative norm>, so multiplication is a <bilinear map> between finite-dimensional <vector spaces>, and hence is smooth.
Let $S=\{x\in A:|x|=1\}$ be the <unit sphere> of the given <positive-definite bilinear form>. It is a copy of $S^{n-1}$. The squaring map
$$
Q:S\to S,\qquad Q(x)=x^2,
$$
is well defined because $|x^2|=|x|^2=1$, and is smooth. <Commutativity> gives its derivative
$$
DQ_x(v)=xv+vx=2xv.
$$
For $x\in S$ and $v\neq0$, the norm identity gives $|DQ_x(v)|=2|v|$, so the derivative is injective, including on the <tangent space> $T_xS=x^\perp$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the <sphere> to itself. Alternatively, polarization of $|xy|^2=|x|^2|y|^2$ gives $\langle xy,xz\rangle=|x|^2\langle y,z\rangle$, so $\langle x^2,2xv\rangle=2\langle x,v\rangle=0$ directly.
The only identifications under squaring are antipodal:
$$
Q(x)=Q(y)\ \Longrightarrow\ (x-y)(x+y)=0\ \Longrightarrow\ y=x\text{ or }y=-x.
$$
The factorization uses <commutativity>, and the last implication uses the absence of <zero divisors>. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth map
$$
q:\mathbb{RP}^{n-1}=S/(x\sim-x)\longrightarrow S^{n-1}.
$$
The antipodal <quotient map> is a local <diffeomorphism>, so the injectivity of $DQ$ implies injectivity of $Dq$. Since <Real projective space> is <compact>, the stated <compact injective immersion is an embedding> theorem makes $q$ a closed embedding.
For $n\geq2$ its source and target have the same dimension $n-1$. The <inverse function theorem> makes its image open as well as closed in the connected <sphere> $S^{n-1}$. It is nonempty, so the image is the whole <sphere>; consequently $q$ is a <diffeomorphism>. But for $n\geq3$ this is impossible:
$$
\pi_1(\mathbb{RP}^{n-1})\cong\mathbb Z/2,
\qquad \pi_1(S^{n-1})=0.
$$
Indeed, the antipodal two-sheeted <covering map> $S^{n-1}\to\mathbb{RP}^{n-1}$ has simply connected total space in this range, and is its <universal cover>, with deck group $\mathbb Z/2$. A <sphere> of dimension at least two is <simply connected>. These <fundamental groups> cannot be those of homeomorphic spaces.
We have obtained the <dimension bound for a commutative Euclidean normed algebra>:
$$
\boxed{\dim_{\mathbb R}A\leq2.}
$$
The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map $\mathbb{RP}^1\to S^1$ is entirely compatible with the argument: the <circle> squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.
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