# Show that there does not exist integers $n_{1}, \ldots, n_{8}$, not necessarily distinct, such that $n_{1}^{4}+\cdots+n_{8}^{4}=1993$.

elementary-number-theory

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.

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.