= Solution
Put $p=4272943$. Since $p\equiv7\pmod8$, the supplementary laws give $(-1/p)=-1$ and $(2/p)=1$. Therefore $(-286/p)=-(143/p)$.
Both $143$ and $p$ are three modulo four. Reciprocity for the <Jacobi symbol>, followed by the supplied remainder, gives
$$
(143/p)=-(p/143)=-(103/143).
$$
Again both denominators are three modulo four, so
$$
(103/143)=-(143/103)=-(40/103).
$$
Now $103\equiv7\pmod8$ and $40=8\cdot5$. Consequently
$$
(40/103)=(2/103)^3(5/103)=(103/5)=(3/5)=-1,
$$
because the nonzero squares modulo five are one and four. Thus $(103/143)=1$, $(143/p)=-1$, and finally
$$
\boxed{(-286/p)=1.}
$$
The congruence \b[is soluble], with exactly two distinct solutions modulo this prime: a nonzero root and its negative. The computation uses a prime denominator in its final step; a positive Jacobi symbol with composite denominator would not by itself imply solubility.
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