Elementary Number Theory exponents

Question

How can one prove that the fifth power of any number has the unit digit same as that of the number itself.

Actually the question was to prove that $n^5-n$ is divisible by $30$ when n is a number prime to 5 and hence prove that the unit digits of the fifth power of a number is equal to the unit digit of number itself.

The first part was easy to be proved, I just took numbers $5m\pm1;5m\pm2$ and proved the rest using binomial theorem(for opening the expression) and mathematical induction. But in the second part I am completely clueless.

Please use only elementary number theory to prove the given statement.

Answer 1

In the second part you need to prove
$n^5 \equiv n (mod 10) $ Since $(2,5)=1$ So just to prove $n^5 \equiv n (mod 2) $ And $n^5 \equiv n (mod 5) $ Which can be proved very easily by induction . Combining these will give the required answer. $$edit$$ : since $30 $ divides $ n^5-n$ i.e. $3×10$ divides $ n^5-n $ or we can say $10 $ divides $ n^5-n $ which gives the direct approach . If you want to prove the second part separately you can do the above proof .

Answer 2

think in terms of the ring $Z_{10}$

the cases $0$ and $1$ are trivial

for the remaining elements in the group of units i.e. $\{3,7,9\}$, we know the group has order 4 so for these elements $x^5 \equiv_{10} x$.

since $2^5 \equiv_{10} 2$, then combining this with the results for $\{3,7,9\}$ we see by multiplicativity that $\{6,4,8\}$ also satisfy the equation

this leaves $5$, which is idempotent.

Answer 3

You have already proved that $n^5 - n$ is divisible by $30$. But by definition this means $n^5 = n + 30k$ for some integer $k$ (which depends on $n$). When you add a multiple of $10$ to a (nonnegative) integer, the last digit doesn't change.

This leaves only the case that $n$ is divisible by $5$, but actually $n^5-n$ is always divisible by $30$ regardless of whether $n$ is relatively prime to $5$, and you can prove this the same way. Or else you can just work out the last digit of $n^5$ when $n$ ends in $0$ or $5$.

Answer 4

Firstly, there's a simpler way to prove the first part than the method mentioned in the question.

Although the question initially asks about $n$ coprime to 5, $30|(n^5 - n)$ is actually true for all $n$, saying that $n$ is coprime to 5 is really a hint that it may be useful to work in modulus 5 at some stage of the proof.

Before we commence we need a lemma:

If $p|n$, $q|n$, with $gcd(p, q) = 1$, then $pq|n$.

This follows from the Fundamental theorem of arithmetic, but let's show it via Euclid's lemma and Bézout's identity.

Since $p|n$ and $q|n$, $\exists\, u, v:$ $$n = pu = qv$$ Also, $\exists\, x, y:$ $$px + qy = 1$$ Multiplying by $n$ and substituting, $$pxqv + qypu = n$$ $$pq(xv + yu) = n$$ Hence $pq|n$

Now we'ready to prove the first part.

Let $f(n) = n^5 - n$. We want to show that $30|f(n)$.

$30 = 2\cdot3\cdot5$ so if we prove that each of 2,3 & 5 divide $f(n)$, then by our lemma $30|f(n)$; in addition we also get $15|f(n)$, $10|f(n)$, and $6|f(n)$ "for free".

Factorising,

$$\begin{align} f(n) & = n^5 - n\\ & = n(n^4 - 1)\\ & = n(n^2 - 1)(n^2 + 1)\\ & = n(n-1)(n+1)(n^2 + 1)\\ \end{align}$$

Exactly one of $\{n-1,\, n,\, n+1\}$ must be divisible by 3, and at least one of them must be even, so by our lemma $6 | (n-1)n(n+1)$ and $6|f(n)$.

[Aside: it is easy to show that any sequence of $m$ contiguous integers is equivalent $\mod m$ to ${k: 1 \le k \le m}$, so the product of that sequence must be divisible by each $k$ in the set and hence is divisible by $m!$.]

Now we need to show that $5|f(n)$. Working mod 5, $n \equiv$ one of $0, \pm1, \pm2$.

Clearly, $n(n-1)(n+1) \equiv 0 \mod 5$ for $n$ congruent to one of $0, \pm1$. For $n\equiv 0$ the 1st factor $\equiv 0$; for $n \equiv 1$, the 2nd factor $\equiv 0$; for $n \equiv -1$, the 3rd factor $\equiv 0$.

Now we need to look at the last factor of $f(n)$, i.e., $n^2+1$.

For $n \equiv \pm2, \, n^2+1 \equiv 4 + 1 \equiv 0$.

So $5|f(n)$ for all $n$, and thus by our lemma, and the previous result, $30|f(n)$.

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Now for the second part.

Since $10|30$, $10|f(n)$ (but our lemma and proof technique in the first part also allows us to conclude $10|f(n)$), so $n^5 - n \equiv 0 \mod 10$ and $n^5 \equiv n \mod 10$.
But $N \mod 10$ is simply the final digit of $N$ written in base 10, so the last digit of $n^5$ is the last digit of $n$