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

• MHB
• anemone
In summary, it is shown that for the greatest positive root $a$ of the equation $x^3 - 3x^2 + 1 = 0$, both $\left\lfloor a^{1788} \right\rfloor$ and $\left\lfloor a^{1988} \right\rfloor$ are divisible by 17. This is due to the fact that the sequence $P_n = a^n + b^n + c^n$ has a period of 16 and is equivalent to 1 mod 17 when n is of the form 8k+4.
anemone
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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.

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

anemone said:
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.

[/sp]

## 1. What is the significance of the numbers 1788 and 1988 in this statement?

The numbers 1788 and 1988 are used to represent two different values of the variable 'a'. In this statement, it is being stated that when a is equal to either 1788 or 1988, the resulting values are both divisible by 17.

## 2. How can it be proven that both ⌊a¹⁷⁸⁸⌋ and ⌊a¹⁹⁸⁸⌋ are divisible by 17?

This statement can be proven by using the property that if a number is divisible by another number, then the ceiling of that number is also divisible by the ceiling of the other number. So if ⌊a¹⁷⁸⁸⌋ and ⌊a¹⁹⁸⁸⌋ are both divisible by 17, then their ceiling values must also be divisible by 17.

## 3. Is there a specific range of values for 'a' that will satisfy this statement?

Yes, there is a specific range of values for 'a' that will satisfy this statement. The values of 'a' must be either 1788 or 1988 in order for the resulting values to be divisible by 17.

## 4. Can this statement be generalized for other numbers besides 17?

Yes, this statement can be generalized for any number that is a factor of 1788 and 1988. For example, if we replace 17 with 2, then the statement would be "⌊a¹⁷⁸⁸⌋ and ⌊a¹⁹⁸⁸⌋ are both divisible by 2".

## 5. Why is this statement important in the field of mathematics?

This statement is important in the field of mathematics because it demonstrates the use of number properties and how they can be used to prove statements. It also highlights the concept of divisibility and how it can be applied to different numbers and values.

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