Can anyone give me hints to how to prove this?? Prove that for any positive integer n, n^5 and n have the same units digit in their base 10 representations; that is, prove that n^5 = n (mod 10). Thanks!
I'm sorry, I'm still a bit lost. Can you please explain what the Euler-Fermat theorem is and how I can apply that to this problem? Thanks
See the following link: http://planetmath.org/encyclopedia/EulerFermatTheorem.html And notice that you only have to prove that: [tex]n^{4}\equiv 1 \left(mod 10\right)[/tex] For n coprime with 10.
This is equivalent to proving that n^{5} - n [itex]\equiv[/itex] 0 (mod 10) You can show this by proving that n^{5} - n is even, and is divisible by 5. The first part is easy, since two of the factors of n^{5} - n are n and n + 1, one of which has to be even for any value of n. The second part, showing that n^{5} - n is divisible by 5 can be done by math induction, and isn't too tricky.