Submitted by jetay1 on Thu, 12/22/2016 - 05:01
What do you think of this way of doing 4(ii) of assignment 17?
We are looking to show that $9\vert(2^n+5^n+56)$. So I looked at $2^n$ and $5^n \pmod{9}$ and saw that they repeated every 6th value of $n$ (in a reverse pattern, interestingly). For $n=0,1,2,\ldots$, $2^n\equiv1,2,4,-1,-2,-4,1,\ldots \pmod{9}$ and $5^n\equiv1,-4,-2,-1,4,2,1\ldots \pmod{9}$.
So for $n=0,1,2,\ldots$, we can work out that $2^n+5^n+56\equiv4,0,4,0,4,0\ldots \pmod{9}$. So 9 divides the expression for all odd $n$ and also every second multiple of 3. And thus 63 divides it (divisibility by 7 having already been shown). And when $n=3$, $2^n+5^n+56\equiv0\pmod{9}$, thus $2^n+5^n+56\equiv0\pmod{9}$ for $n\bmod3=0$ and odd $n$.
It's not nearly as elegant as the idea of setting $n=3m$, but it seems logically sound.
PS There are a couple of typing mistakes in the hints sheet for this question. $5^n$ has become $5^2$ in a few places, and there's a $1$ missing from $2^{3m+}$ in the blue equation.
Noting that $2^n$ cycles
Noting that $2^n$ cycles through 6 values in reverse order to how $5^n$ cycles through them is a good *observation*, but it's not quite a proof, you've looked at the first few values and saw a pattern, but you need to show this pattern actually exists.
(i) It's periodic. $2^{n+6} \equiv 2^n \cdot 2^6 \equiv 2^n \pmod{9}$ since $2^6 = 64 = 63 + 1 \equiv 1 \pmod{9}$. (you can do the same thing for $5^n$, if you wish)
(ii) Using somewhat the same idea, can you now show it cycles in reverse order to $5^n$?
Mistakes
Thanks for pointing out the mistakes in the hints sheet. I'll fix them after Christmas!
Mistakes fixed. Hopefully no
Mistakes fixed. Hopefully no new ones added!
Thanks
Thanks very much. That's very helpful advice.