<s>[INST] 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?
 [/INST] 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 $n=1$.
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