Explain why n^5 - n is divisible by 5 for n=0,1,2,3,....
Can anyone explain to me how to do this?(without mod - I haven't learnt that)
Thanx
Hello,
$\displaystyle n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$
If $\displaystyle n=5k$ : it is divisible by 5.
If $\displaystyle n=5k+1 \implies n-1=5k$ : it is divisible by 5.
If $\displaystyle n=5k+4 \implies n+1=5k+5=5(k+1)$ : it is divisible by 5.
If $\displaystyle n=5k+2 \implies n^2+1=(25k^2+20k+4)+1=5(5k^2+4k+1)$ : it is divisible by 5.
If $\displaystyle n=5k+3 \implies n^2+1=(25k^2+30k+9)+1=25k^2+30k+10=5(5k^2+6k+2)$ : it is divisible by 5.
You (or rather I...) have done all the possible situations.
yes it is.Originally Posted by Showcase_22
a is divisible by b if when a is divided by b, the quotient is an integer (0 is an integer) and the remainder is 0.
Moo is on the right track. However, note that
$\displaystyle n^5-n=n(n+1)(n-1)(n^2+1)$
$\displaystyle \color{white}.\hspace{10mm}.$ $\displaystyle =n(n+1)(n-1)(n^2-4+5)$
$\displaystyle \color{white}.\hspace{10mm}.$ $\displaystyle =n(n+1)(n-1)(n^2-4)+5n(n+1)(n-1)$
Now $\displaystyle 5n(n+1)(n-1)$ is divisible by 5, while $\displaystyle n(n+1)(n-1)(n^2-4)=(n-2)(n-1)n(n+1)(n+2)$ is the product of 5 consecutive integers and so is also divisible by 5; hence their sum is divisible by 5.
You did not show that $\displaystyle (k-5)(k+1)(k+2)(k^2+2k+2)$ is divisible by 5 for all $\displaystyle k$, only that it’s divisible by 5 for $\displaystyle k=5.$ That’s not good enough.
If you want to use induction, there is no need to factorize. Just use $\displaystyle (n+1)^5-(n+1)=(n^5-n)+5(n^4+2n^3+2n^2+n).$
I thought that you assumed it was true for n=k+1 and you could only say it is true after you've proven it?
In this case, I assumed it was true and worked through it. If it wasn't true I wouldn't have ended up with something that was divisible by 5 at the end.
You don’t assume it is true for $\displaystyle n=k+1$. In proof by induction, you assume your formula is true for $\displaystyle n=k$, then prove it is true for $\displaystyle n=k+1$.
Be careful what you can and cannot assume in proving something. In proof by contradiction, you assume that the result you want to prove is false, then show that this leads to a contradiction. However, you never assume what you want to prove is already true before proving it. This fallacy is called “begging the question”.
ahhhh, I get it!
So I was begging the question when I assumed it was true for n=k+1 ie. I assumed it was true before I had proven it.However, you never assume what you want to prove is already true before proving it. This fallacy is called “begging the question”.
Well I won't make tha mistake again. Thankyou!