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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

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    $\begingroup$ but you need atleast two of them to be consecutive $\endgroup$
    – Tyrone
    Commented Aug 12, 2019 at 18:17
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    $\begingroup$ Hint: Any number raised to the 4th power is either 0 or 1 mod 16, depending on whether the number you start with is even or odd. $\endgroup$
    – Nate
    Commented Aug 12, 2019 at 18:17
  • $\begingroup$ Sorry, missed that condition. $\endgroup$
    – Thomas Andrews
    Commented Aug 12, 2019 at 18:18

2 Answers 2

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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 $n_i$.

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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

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    $\begingroup$ Well, what the note means is that either all $n_i$ are even or all $n_i$ are odd. You're argument is close, but you need that if they are consecutive, then one is even and one is odd. Knowing that one number is even is not enough. $\endgroup$
    – Thomas Andrews
    Commented Aug 12, 2019 at 18:29

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