Show ⌊a¹⁷⁸⁸⌋ and ⌊a¹⁹⁸⁸⌋ are both divisible by 17

[anemone]

Show ⌊a¹⁷⁸⁸⌋ and ⌊a¹⁹⁸⁸⌋ are both divisible by 17

Let $a$ be the greatest positive root of the equation $x^3− 3x^2+ 1 = 0$.

Show that $\left\lfloor{a^{1788}}\right\rfloor$ and $\left\lfloor{a^{1988}}\right\rfloor$ are both divisible by 17.

 

[Opalg]

Re: Show ⌊a¹⁷⁸⁸⌋ and ⌊a¹⁹⁸⁸⌋ are both divisible by 17

Let $a$ be the greatest positive root of the equation $x^3− 3x^2+ 1 = 0$.

Show that $\left\lfloor{a^{1788}}\right\rfloor$ and $\left\lfloor{a^{1988}}\right\rfloor$ are both divisible by 17.

[sp]Let $f(x) = x^3− 3x^2+ 1$. Then $f(-1) = -3$, $f(0) = 1$, $f(1) = -1$, $f(2) = -3$ and $f(3) = 1$. By the intermediate value theorem, the largest root, $a$, lies between $2$ and $3$; and the other two roots, $b$ and $c$, lie between $-1$ and $1$.

Let $P_n = a^n + b^n + c^n$. By Newton's identities, $$P_0 = P_1 = 3,\quad P_2 = 9, \quad P_3 = 24, \quad P_n = 3P_{n-1} - P_{n-3}\ \ (n>3).$$ Working mod $17$, it is easy to use that recurrence relation to calculate $P_n \pmod{17}$: $$ \begin{array}{r|ccccccccccccccccccc}n& 0&1&2&3&4 & 5&6&7&8&9 & 10&11&12&13&14 & 15&16&17&18 \\ \hline P_n\pmod{17}& 3&3&9&7&1 & 11&9&9&16&5 & 6&2&1&14&6 & 0&3&3&9 \end{array}.$$ Notice that the sequence starts to repeat when $n=16$, and therefore has period $16$. Also, $P_4 = P_{12} = 1 \pmod{17}$. It follows that $P_n = 1 \pmod{17}$ whenever $n$ is of the form $8k+4$.

But $P_n = a^n + b^n + c^n$. When $n$ is large and even, $b^n$ and $c^n$ will both be small and positive. In particular, $0<b^n+c^n<1$. So if $P_n=1\pmod{17}$ then $\left\lfloor a^n \right\rfloor = 0\pmod{17}$. Since $1788$ and $1988$ are both of the form $8k+4$, the result is proved.

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