While teaching binomial expansion, one of my high school students asked me the following question:
What is the last digit of $(1+\sqrt{71})^{71}+(1-\sqrt{71})^{71}$?
I have absolutely no context on how he came up with it, but it seemed interesting enough.
$(1+\sqrt{71})^{71} + (\sqrt{71}-1)^{71}$ is not an integer.
However, if you meant $(1+\sqrt{71})^{71} + (1-\sqrt{71})^{71}$, then it is indeed an integer.
If $x_n = (1+\sqrt{71})^n + (1-\sqrt{71})^n$, we then have $x_n$ to satisfy the recurrence $$x_{n+1} = 2x_n + 70x_{n-1}$$ where $x_0 = 2$ and $x_1=2$. Hence, we see that $$x_{n+1} = 2x_n \pmod{10}$$ Since $x_1 \equiv 2\pmod{10}$, we see that $$x_n \equiv 2^n \pmod{10}$$ which implies $$x_{4k} \equiv 6\pmod{10}$$ $$x_{4k+1} \equiv 2\pmod{10}$$ $$x_{4k+2} \equiv 4\pmod{10}$$ $$x_{4k+3} \equiv 8\pmod{10}$$ Since $71 \equiv3\pmod4$, we obtain that $x_{71} \equiv 8 \pmod{10}$, i.e., the last digit is $8$.