Show that there does not exist integers $n_{1}, \ldots, n_{8}$, not necessarily distinct, such that $n_{1}^{4}+\cdots+n_{8}^{4}=1993$. I tried using mod of $8$ and got that mod $8$  of an even number raised to the power $4$ is $0$, and $1$ in case of odd number. Also, $1993$ is $1$ mod $8$, but can't make progress from this point. Also, can't figure out what to do with the information that they aren't necessarily distinct.
 A: You have the right idea, but using modulo $16$ instead works better (as also suggested in Arturo Magidin's comment). Similar to modulo $8$, in modulo $16$, every even number raised to the power $4$ is $0$, and is $1$ for odd numbers. This means the left side sum can only be a value between $0$ and $8$, inclusive, modulo $16$. However, $1993 \equiv 9 \pmod{16}$, which proves there are no integers $n_{1}, \ldots, n_{8}$ that solve the equation.
As for these integers not necessarily being distinct, the requirement doesn't make any difference with this proof. I suspect it's likely mentioned in your problem just to keep it as general as possible (i.e., so we can't exclude possible solutions just because any of the $x_i$ values are repeated) in case a different approach is attempted instead, e.g., if some form of brute force was tried. Also, if they were required to be distinct instead, then since $8^4 = 4096$, we have $|n_i| \le 6$. However, using modulo $8$ as you did, there needs to be $7$ even integers among the $n_i$, but $n_i \in \{-6, -4, -2, 0, 2, 4, 6\}$ which has $7$ elements. Thus, each integer in that set must be used exactly once, but $(-6)^4 + 6^4 = 2592$, so there would be no solutions.
A: HINT.-You have $n^4\equiv0\pmod{16}$ when $n$ is even and $n^4\equiv1\pmod{16}$ when $n$ is odd then taking modulo $16$ you have $ n_{1}^{4}+\cdots+n_{8}^{4}\le8$.
But $1993\equiv 9\pmod{16}$
