Figure out all positive integers n with consecutive + integers a,b,c. When $2018^n$ = $a^4$ + $b^4$ + $({b^2+c^2})^2$,
then what is the possible positive integers n be?

Let us consider the given equation modulo $4$. Then $a,b$ are consecutive, of the shape $2k$ and $2k\pm1$ (in the right order), so $a^4+b^4$ is $0+1$ or $1+0$ modulo four. The same argument shows $b^2+c^2\equiv 0+1$ (or $1+0$) modulo four. The whole expression is thus $1+1^2=2$ modulo four.
So on the L.H.S of the given equation we have exactly the first power, since $1094\equiv 2$ modulo four. Sad enough, there is no solution, we compute for instance in sage:
sage: R.<x> = PolynomialRing(ZZ)
sage: a, b, c = x-1, x, x+1
sage: factor( a^4 + b^4 + (b^2+c^2)^2 - 1094 )
2 * (3*x^4 + 2*x^3 + 7*x^2 - 546)

and there is no linear factor.
