Assignment 17

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

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