The remainder left out when $8^{2n}-62^{2n+1}$ is divided by $9$ is?

Question

I was solving this question while practicing binomial theorem, and my $1^{st}$ thought was to substitute $n=0$ for which I get $-61$ and find the remainder which I get as $7$. And when I checked the answer it was $2$. So I tried substituting $0$ again to check my answer, and I got $7$ again, as the remainder. I substituted $n=1$ and still got $7$ as the remainder. I am really confused by the book's solution too. Can someone please say where I made the mistake?
Book answer:
$$8^{2n}-62^{2n+1}= \ (9-1)^{2n}- \ (63-1)^{2n+1}$$ $$= (\ ^{2n}C_{0}9^{2n}-^{2n}C_{1}9^{2n-1}+ \ ... \ +^{2n}C_{2n}9^{0}) \ - \ (^{2n+1}C_{0}63^{2n}+ \ ... \ ^{2n+1}C_{2n+1}63^0)$$ $$=9(k_1+7k_2)+2$$ $$=9\beta + 2 \ \ \ (\text{where $k_1$,$k_2$ and $\beta$ are some integers.})$$

Answer 1

For $n=0$ we have $$1-62^{1}=-61=9\cdot(-7)+2$$ where $\beta=-7$, so the remainder is $2.$

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The Division Theorem tell us that for every pair of integers $a,b$ where $b\neq 0$, there exist unique integers $q,r$ such that $$a=qb+r$$ and $0\leq r< |b|$. Comparing above we have $a=-61$ and $b=9$ which give $q=-7$ and $r=2$ with $0\leq r<9$.

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From the book's solution $$ (\ ^{2n}C_{0}9^{2n}-^{2n}C_{1}9^{2n-1}+ \ ... \ +\color{red}{^{2n}C_{2n}9^{0}}) \ - \ (^{2n+1}C_{0}63^{2n}+ \ ... \ -\color{\red}{^{2n+1}C_{2n+1}63^0})$$

All the non-red terms above are divisible by $9$, moreover since $63=9\times 7$, we can pull a factor of $7$ from the second bracket after excluding the red term. Since the two red terms are $1$ we can re-write it as $$ (\ ^{2n}C_{0}9^{2n}-^{2n}C_{1}9^{2n-1}+ \ ... \ ) \ - \ (^{2n+1}C_{0}63^{2n}+ \ ... \ )+\color{red}{^{2n}C_{2n}9^{0}}+\color{\red}{^{2n+1}C_{2n+1}63^0}$$ $$=(\ ^{2n}C_{0}9^{2n}-^{2n}C_{1}9^{2n-1}+ \ ... \ ) \ - \ \color{blue}{7}(^{2n+1}C_{0}9^{2n}7^{2n-1}+ \ ... \ )+2$$ $$=9(k_{1}+7k_{2})+2$$ $$=9\beta+2$$

where $k_1$,$k_2$ and $\beta$ are some integers as required.

Answer 2

Your technique and answer are correct, but it can be made easier. Building on @MathLover's comment, work modulo $9$ so $a\equiv b\implies a^n\equiv b^n$, because $a-b|a^n-b^n$. So$$8\equiv62\equiv-1\implies8^{2n}-62^{2n+1}\equiv(-1)^{2n}-(-1)^{2n+1}=2(-1)^{2n}=2.$$

Answer 3

Alternatively, let $f(n)=8^{2n}-62^{2n+1}$. Then $f(n+1)-f(n)=63 (8^{2n} - 61 \cdot 62^{2 n + 1})$ is a multiple of $9$. Therefore, the remainder of $f(n)$ when divided by $9$ is the same for all $n$. Since $f(0)=-61$, the remainder is always $2$.