I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$. However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!
[Math] $n^5-n$ is divisible by $10$
algebra-precalculusdivisibilityelementary-number-theory
Related Question
- [Math] Find the last $4$ digits of $2016^{2016}$
- [Math] A positive integer (in decimal notation) is divisible by 11 $ \iff $ …
- [Math] If $p \not = 5$ is an odd prime, prove that either $p^2+1$ or $p^2-1$ is divisible by $10$
- Does there exist a composite, deficient, odd number that is divisible by the sum of its proper factors
- The next to last digit of the square of a natural number is odd. Prove that its last digit is $6$ .
Best Answer
Your proof is good enough. There's a slight improvement, if you want to avoid modular arithmetic / considering cases.
$n^5 - n$ is a multiple of 5 $\Leftrightarrow$ $ n^5 + 10 n^4 + 35n^3 + 50 n^2 + 24 n = n^5 -n + 5(2n^4 + 7n^3 + 10n^2 + 5n) $ is a multiple of 5. The latter is just $n(n+1)(n+2)(n+3)(n+4)$, which is the product of 5 consecutive integers, hence is a multiple of 5.
Note: You should generally be able to do the above transformation, and can take the product of any 5 (or k) consecutive integers, if you are looking at a polynomial of degree 5 (or k).