Property of Integer Raised to the Tenth Power

Question

Let $x$ be an integer. I want to show that $x^{10} \in \{-1, 0, 1\} (\mathrm{mod \; 25})$. I can focus on the case where $x$ is positive since we're raising it to an even power. I tried to do this by induction, and you get three cases for the inductive step. I could only do the case where the previous integer was a multiple of 25. Assume $n^{10} \equiv 0(\mathrm{mod} \; 25)$. We have $$(n+1)^{10} \equiv \sum_{k=0}^{10}{10 \choose k}n^{10-k}1^k (\mathrm{mod} \; 25)$$ Since $n^{10}$ is divisible by 25 then $n = 5^\alpha m$, where $\alpha \geq 1$ and $m$ is a positive integer. In that case $$\sum_{k=0}^{10}{10 \choose k}n^{10-k}1^k \equiv \sum_{k=1}^{10} {10 \choose k} (2(5^{\alpha})m)^{10-k} \equiv 1 (\mathrm{mod} \; 25)$$ I do not know how to deal with the cases where $n$ is one more or one less than a multiple of 25, however. Any help?

Answer 1

By looking at last digit of $x$ you can see that $x^2$ can end with $0,1,4,5,6$ or $9$ and thus $5\mid x^2\pm 1$ or $5\mid x$.

• Clearly if $5\mid x$ then $25\mid x^{10}$.
• Say $5\nmid x$, then $x^2=5y\pm 1$ so by binomial theorem we have: \begin{align}x^{10}&=(5y\pm 1)^5\\&= 5^5y^5\pm 5^5y^4 + 2\cdot 5^4y^3 \pm 2\cdot 5^3y^2+5^2y\pm1 \end{align} which means $x^{10} \equiv \pm 1 \pmod {25}$