= Solution
Take a standard <Brownian motion> $B$ and define
$$
\boxed{M_t=B_{t\wedge1},\qquad N_t=B_{t\wedge1}^2-(t\wedge1)}.
$$
The <Itô formula> gives $N_t=2\int_0^{t\wedge1}B_s\,dB_s$, so both are <continuous martingales> starting at zero. Their second moments are $\mathbb EM_t^2=t\wedge1$ and $\mathbb EN_t^2=2(t\wedge1)^2$, making them <L2-bounded continuous martingales>. At equal times, the odd moments of a centered <normal random variable> vanish, so
$$
\mathbb E[M_tN_t]=\mathbb E[B_{t\wedge1}^3]-(t\wedge1)\mathbb EB_{t\wedge1}=0.
$$
They are consequently <weakly orthogonal continuous martingales> by <a/i/solution>.
Their product, for $t<1$, is $B_t^3-tB_t$. The <Itô formula> gives
$$
d(M_tN_t)=(3B_t^2-t)\,dB_t+2B_t\,dt.
$$
The continuous finite-variation term $2\int_0^tB_s\,ds$ is not identically zero: it has variance $4t^3/3$ at every $t>0$. Uniqueness of the continuous semimartingale decomposition therefore shows that $MN$ is not even a <local martingale>. Equivalently $[M,N]_t=2\int_0^{t\wedge1}B_s\,ds$ is not zero, so the pair is not <orthogonal continuous local martingales>.
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