So I want to prove that is in title. I'm working with modular arithmetic, so if any of you can show me a hint, it will be perfect.
Asked
Active
Viewed 183 times
0
-
Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) \bmod 64$. – lhf Jan 26 '19 at 01:19
-
1@lhf Easier to use $\ x_{n+1}-\color{#c00}9,x_n = 64(4-5n).\ $ This works generally - see this answer in the dupe. – Bill Dubuque Jan 26 '19 at 03:24
2 Answers
3
We find that $$ \frac{3^{2n+3}+40n-27}{8}=\frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+\cdots+1)+5n. $$ And also $$ 27(9^{n-1}+9^{n-2}+\cdots+1)+5n\equiv 27(1+1+\cdots +1)+5n\equiv32n\equiv 0\ \;\text{(mod}\;8). $$ Therefore $64 \;|\; 3^{2n+3}+40n-27$ for all $n\ge 1$.
Myunghyun Song
- 21,723
- 2
- 24
- 60
-
i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ? – Richard Kingston Jan 26 '19 at 00:10
-
@RichardKingston Basically, yes. If $8\ | \ N$ and $8\ | \ (N/8)$, then it holds $64 \ | \ N$ . – Myunghyun Song Jan 26 '19 at 00:12
0
Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27\equiv 0\mod64$. We have to show that $$3^{2n+5}+40(n+1)-27=9\cdot3^{2n+3}+40n+13\equiv 0\mod64.$$ The inductive hypothesis can be rewritten as $$3^{2n+3}\equiv -40n +27\mod64,$$ so that $$9\cdot3^{2n+3}+40n+13\equiv-360n+243+40m+13=-320n+256\equiv 0\mod 64. $$
Bernard
- 175,478