Hints for Proving n^5 = n (mod 10)

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To prove that n^5 ≡ n (mod 10) for any positive integer n, it is essential to demonstrate that n^5 - n is divisible by both 2 and 5. The evenness of n^5 - n follows from the factors n and n + 1, ensuring at least one is even. For divisibility by 5, mathematical induction can be employed to show that n^5 - n is indeed divisible by 5 for all positive integers n. Additionally, applying the Euler-Fermat theorem indicates that n^4 ≡ 1 (mod 10) for n coprime with 10, which supports the proof. This establishes that n^5 and n share the same units digit in their base 10 representations.
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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!
 
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What does the Euler-Fermat theorem tells you, when applied to a congruence mod 10?
 
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:

n^{4}\equiv 1 \left(mod 10\right)

For n coprime with 10.
 
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lil_luc said:
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!
This is equivalent to proving that n5 - n \equiv 0 (mod 10)

You can show this by proving that n5 - n is even, and is divisible by 5.
The first part is easy, since two of the factors of n5 - n are n and n + 1, one of which has to be even for any value of n.
The second part, showing that n5 - n is divisible by 5 can be done by math induction, and isn't too tricky.
 
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