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Studying GCD, I got a question that begs to show that $n$ and $n^5$ has the same units digit ... What would be an idea to be able to initiate such a statement? testing $0$ and $0^5=0$ $1$ and $1^5=1$ $2$ and $2^5=32$

In my studies, I have not got "mod", please use other means, if possible of course.

I demonstrated in a previous period that $$2|n^5-n$$because $$n^5-n=(n+1)n(5n^4+5n+5)$$, and$$5|n^5-n$$By Fermat's Little Theorem

Only I do not understand what should happen to the units of the two numbers are equal ... What must occur?

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1  
Well, $n\equiv n^5\pmod2$ and $n\equiv n^5\pmod5$. –  awllower Jul 25 '13 at 14:48
    
Although not studied'' mod''. Using GCD has no like? –  marcelolpjunior Jul 25 '13 at 14:52
    
Well, mod is gcd, since $a\equiv b\pmod n$ if and only if $\gcd(n,a-b)=n$. –  awllower Jul 25 '13 at 14:56

4 Answers 4

up vote 7 down vote accepted

Without using any modular arithmetic:

$$n^5-n=n(n-1)(n+1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)=n(n-1)(n+1)(n^2-4)+5n(n-1)(n+1)=$$ $$=(n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1)$$

$(n-2)(n-1)n(n+1)(n+2)$ is the product of 5 consecutive integers thus divisible by 2 and 5.

$5n(n-1)(n+1)$ is multiple of $5$ and even.

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We know if unit digits of two numbers are same, their difference is divisible by 10 and vice versa.

Method $1a:$

Using Fermat's Little Theorem $n^5-n\equiv0\pmod 5$

and $n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$ which is divisible by $n(n-1)$ which is always even

$\implies 2|(n^5-n)$ and we have $5|(n^5-n)$

$\implies n^5-n$ is divisible by lcm $(2,5)=10$

Method $1b:$

As $10=2\cdot5,$

using Fermat's Little Theorem, we have $$n^5-n\equiv0\pmod 5\text{ and } n^2-n\equiv0\pmod 2$$

Now, lcm $(n^5-n,n^2-n)=n(n^4-1,n-1)=n(n^4-1)$ as $(n-1)|(n^4-1)$

$\implies $lcm $(n^5-n,n^2-n)=n^5-n$ which is divisible by $5,2$ hence by lcm$(2,5)=10$


Method $2:$

Alternatively, $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$ $$=n(n^2-1)(n^2-4)+5n(n^2-1)$$ $$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\cdot \underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$

Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer

So, $(n-2)(n-1)n(n+1)(n+2)$ is divisible by $5!=120$ and $(n-1)n(n+1)$ is divisible by $3!=6$

$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$

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Damn, too fast! –  Arkamis Jul 25 '13 at 14:48
    
Or $n$ and $n^5$ are both even or both odd ... –  Mark Bennet Jul 25 '13 at 14:48
    
@lab bhattacharjee: Is there another way? Without using mod. –  marcelolpjunior Jul 25 '13 at 14:54
    
@MarkBennet Then either $n$ is of the form $2k$ or $2k +1$ order. Right? –  marcelolpjunior Jul 25 '13 at 14:55
    
@marcelolpjunior, I am adding an alternative solution –  lab bhattacharjee Jul 25 '13 at 14:57

The result can be verified using minimal machinery. first note that the units digit of $n^5$ is completely determined by the units digit of $n$.

The units digit of $n^5$ is one of $0,1,2,\dots,9$. We can verify the result for each of the $10$ cases by a direct calculation.

There are $10$ calculations to do, none of them painful. We do one of them. Let $n$ end in $8$. Then $n^2$ ends in $4$, so $n^4$ ends in $6$. Thus $n^5$ ends in $8$.

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You can write $n=10a+b$, with $b$ the units digit. Then $n^5=(10a)^5+5(10a)^4b+10(10a)^3b^2+10(10a)^2b^3+5(10a)b^4+b^5$. As all the terms but the last have a factor of $10$, the units digit of $n^5$ is the same as the units digit of $b^5$. This justifies testing each units digit to see if the units digit of its fifth power is the same as the digit. Ten tests and you are done-they all succeed.

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1  
Actually 6 tests are enough, since you can write $n=10a \pm b$ where $b$ is a digit $\leq 5$. –  N. S. Jul 25 '13 at 15:08
    
@N.S.: Good point. –  Ross Millikan Jul 25 '13 at 15:11

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