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Is it possible to prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ for $n \geq 1 $ . I know that it is true because $ \frac{3^n - 4(2^n) + (-1)^n + 6}{24}$ represents the number of ways to uniquely $4$-colour an n-cycle , excluding permutations of colours.

A. Napster
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3 Answers3

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Let us try modulo $\pmod3,\pmod8$ separately.

As $4\equiv1\pmod3,2\equiv-1\implies2^n\equiv(-1)^n$

$$3^n-4\cdot2^n+(-1)^n+6\equiv-(-1)^n+(-1)^n\equiv0\pmod3$$

For $n\ge1,8\mid4\cdot2^n$

$$3^n-4\cdot2^n+(-1)^n+6\equiv3^n+(-1)^n+6\pmod8$$

Now as $3^2\equiv1\pmod8,(3^2)^m\equiv1$

If $n$ is even, $$3^n+(-1)^n+6\equiv9^m+1+6\pmod8\equiv1+1+6$$

What if $n$ is odd $=2m+1$

lab bhattacharjee
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Using weak induction,

if $f(n)=3^n - 4(2^n) + (-1)^n + 6,$

$$f(m+2)-f(m)=3^m(3^2-1)-4\cdot2^m(2^2-1)$$ which is clearly divisible by $3\cdot8$ for $m\ge1$

So, $24\mid f(m)\iff24\mid f(m+2)$

Now establish the base cases $f(1),f(2)$

lab bhattacharjee
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  • $$f(m+2)+f(m+1)=3^{m+1}(3+1)-4\cdot2^{m+1}(2+1)+12=12(3^m+1)-12\cdot2^{m+1}$$ As $3^m+1$ is even for $m\ge0,$ $24$ divides $$f(m+2)+f(m+1)\implies24|f(m+2)\iff24|f(m+1)$$ Now establish the base case $f(1)$ – lab bhattacharjee Dec 19 '17 at 13:51
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Let $x_n = 3^n - 4(2^n) + (-1)^n$. By expanding $(x-3)(x-2)(x+1)=x^3 - 4 x^2 + x + 6$, we get $$ x_n= 4x_{n-1}-x_{n-2}-6x_{n-3} \quad\mbox{ for } n\ge 4 $$ Let $y_n = x_n + 6$. Then $$ y_n= 4y_{n-1}-y_{n-2}-6y_{n-3} + 24 $$ and the claim follows immediately by induction after checking it for $y_1, y_2, y_3$, which are all zero.

lhf
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  • This is still induction but at least it is not blind induction and explains where $24$ comes from. – lhf Dec 19 '17 at 14:02