$\{a\}+\{1/a\}=1$ prove that $\{a^3\}+\{1/a^3\}=1$

Question

If $a$ is a real number such that If $\{a\}+\{1/a\}=1$ prove that $\{a^3\}+\{1/a^3\}=1$, where $\{x\}=x-[x]$ for any $x$ real number I got that $a+1/a-[a]-[1/a]=1$ but I do not think this is useful in some way.

Answer 1

Hint: Show that $ \{a^3\} + \{ 1/a^3 \}$ is an integer that is strictly between 0 and 2, hence it is 1.

- Show that $ \{ a \} \neq 0$.
- Show that $ 0 < \{a\}, \{ 1/a \} < 1$.
- Show that $ 0 < \{ a^3 \}, \{ 1/ a^3 \} < 1$.
- Show that $ 0 < \{ a^3 \} + \{ 1/ a^3 \} < 2$.

- Show that $ a +1/a$ is an integer. (EG Using the "think not useful" equation you wrote)
- Show that $ a^3 + 1/a^3$ is an integer.
- Show that $ \{ a^3 \} + \{ 1/a^3 \} = a^3 + 1/a^3 - \lfloor a^3 \rfloor - \lfloor 1/a^3 \rfloor $ is an integer.

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Notes

• The brute force approach of solving $ a + 1/a = n$ using the quadratic forumla, and then substituting into the expression, also works.