How many solution are there to the equation $n_1^4+n_2^4+...+n_{16}^4=65536$ with non-negative integers ($n_1,n_2,...,n_{16}$), of which at least two are consecutive?
I know $65536=2^{16}=16^4$ but I cant find any solutions and I don't know how to prove there aren't any solutions
solutions, suggestions and hints would all be appreciated
from the 2018 South African Senior Team Competition http://www.samf.ac.za/content/files/QuestionPapers/2018_Senior_Team_COMBO.pdf
On
as Nate pointed out every number raised to the 4th power is either 0 mod 16 or 1 mod 16 but if two numbers are consecutive that means atleast one of those numbers will be even resulting in 0 mod 16 when raised to the 4th power, which means the LHS can only reach a maximum of 15 mod 16 but the RHS is 0 mod 16 , therefore there are no solutions
Nate's hint (that $n^4 \equiv 0,1 \mod 16$ depending on whether $n$ is even or odd, respectively) gives you the answer. Since $65536\equiv 0 \mod 16$, if any of the $n_i$ are odd, then all of the $n_i$ must be odd so that their sum is $16\equiv 0 \mod 16$. But there are no two consecutive odd numbers. If you then look at all $n_i$ being even, you have the same problem. So there is no solution containing two consecutive values of $n_i$.