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?
 A: 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.
A: Since we have consecutive integers $a,b,c$, take $a=b-1$ and $c=b+1$. Then, we have $2^n 547^n=(b-1)^4+b^4+(b^2+(b+1)^2)^2=(b-1)^4+2b^4+2b^2(b+1)^2+(b+1)^4=(b-1)^4+(b+1)^4+4b^4+4b^3+2b^2$.  If $b=2k$, then expanding the braces, we get the RHS is divisible by $2$, not $4$. This means $n$ is at most 1.
If $b=2k+1$, then again the RHS is divisible by 2, not 4. So $n \leq 1$.
We check the case $n=1$. If $b=4$, the RHS is $2018$. So, we conclude that $n=1$.
