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Q1, Q2: Such a representation exists for all odd powers of $\phi$ as we can show by induction. Using the arctangent identity, let us first write:

$\arctan \phi^{2n+1} = \arctan\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v} - \arctan v$

Next, we set the argument $\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}$ equal to $\phi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\phi$, we end up with $v = \frac{\phi^{2n-1}-\phi^{2n+1}}{1+\phi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $\arctan\left(-\frac{1}{x}\right) = -(2\arctan 1-\arctan x)$, this proves the claim for odd powers of $\phi$.

Q3: For even powers if we set up the analogous equation:

$\arctan \phi^{2n} = \arctan\frac{\phi^{2n}+v}{1-\phi^{2n}v} - \arctan v$

However in this case, setting $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $\arctan\phi^{2n}$ in terms of $\arctan\phi^{2n-2}$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^k$ for other powers $k$ such as $k=2n-1$, so if any $\arctan \phi^{2n}$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.

Q1, Q2: Such a representation exists for all odd powers of $\varphi$ as we can show by induction. Using the arctangent identity, let us first write:

$\arctan \varphi^{2n+1} = \arctan\frac{\varphi^{2n+1}+v}{1-\varphi^{2n+1}v} - \arctan v$

Next, we set the argument $\frac{\varphi^{2n+1}+v}{1-\varphi^{2n+1}v}$ equal to $\varphi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\varphi$, we end up with $v = \frac{\varphi^{2n-1}-\varphi^{2n+1}}{1+\varphi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $\arctan\left(-\frac{1}{x}\right) = -(2\arctan 1-\arctan x)$, this proves the claim for odd powers of $\varphi$.

Q3: For even powers if we set up the analogous equation:

$\arctan \varphi^{2n} = \arctan\frac{\varphi^{2n}+v}{1-\varphi^{2n}v} - \arctan v$

However in this case, setting $\frac{\varphi^{2n}+v}{1-\varphi^{2n}v} = \varphi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $\arctan\varphi^{2n}$ in terms of $\arctan\varphi^{2n-2}$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\varphi^{2n}+v}{1-\varphi^{2n}v} = \varphi^k$ for other powers $k$ such as $k=2n-1$, so if any $\arctan \varphi^{2n}$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.

Q1, Q2: Such a representation exists for all odd powers of $\phi$ as we can show by induction. Using the arctangent identity, let us first write:

$arctan(\phi^{2n+1}) = arctan(\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}) - arctan(v)$

Next, we set the argument $\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}$ equal to $\phi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\phi$, we end up with $v = \frac{\phi^{2n-1}-\phi^{2n+1}}{1+\phi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $arctan(-\frac{1}{x}) = -(2\;arctan(1)-arctan(x))$, this proves the claim for odd powers of $\phi$.

Q3: For even powers if we set up the analogous equation:

$arctan(\phi^{2n}) = arctan(\frac{\phi^{2n}+v}{1-\phi^{2n}v}) - arctan(v)$

However in this case, setting $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $arctan(\phi^{2n})$ in terms of $arctan(\phi^{2n-2})$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^k$ for other powers $k$ such as $k=2n-1$, so if any $arctan(\phi^{2n})$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.

Q1, Q2: Such a representation exists for all odd powers of $\phi$ as we can show by induction. Using the arctangent identity, let us first write:

$\arctan \phi^{2n+1} = \arctan\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v} - \arctan v$

Next, we set the argument $\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}$ equal to $\phi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\phi$, we end up with $v = \frac{\phi^{2n-1}-\phi^{2n+1}}{1+\phi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $\arctan\left(-\frac{1}{x}\right) = -(2\arctan 1-\arctan x)$, this proves the claim for odd powers of $\phi$.

Q3: For even powers if we set up the analogous equation:

$\arctan \phi^{2n} = \arctan\frac{\phi^{2n}+v}{1-\phi^{2n}v} - \arctan v$

However in this case, setting $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $\arctan\phi^{2n}$ in terms of $\arctan\phi^{2n-2}$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^k$ for other powers $k$ such as $k=2n-1$, so if any $\arctan \phi^{2n}$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.

Q1, Q2: Such a representation exists for all odd powers of $\phi$ as we can show by induction. Using the arctangent identity, let us first write:

$arctan(\phi^{2n+1}) = arctan(\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}) - arctan(v)$

Next, we set the argument $\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}$ equal to $\phi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\phi$, we end up with $v = \frac{\phi^{2n-1}-\phi^{2n+1}}{1+\phi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $arctan(-\frac{1}{x}) = -(\frac{1}{2}arctan(1)-arctan(x))$, this proves the claim for odd powers of $\phi$.

Q3: For even powers if we set up the analogous equation:

$arctan(\phi^{2n}) = arctan(\frac{\phi^{2n}+v}{1-\phi^{2n}v}) - arctan(v)$

However in this case, setting $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $arctan(\phi^{2n})$ in terms of $arctan(\phi^{2n-2})$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^k$ for other powers $k$ such as $k=2n-1$, so if any $arctan(\phi^{2n})$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.

Q1, Q2: Such a representation exists for all odd powers of $\phi$ as we can show by induction. Using the arctangent identity, let us first write:

$arctan(\phi^{2n+1}) = arctan(\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}) - arctan(v)$

Next, we set the argument $\frac{\phi^{2n+1}+v}{1-\phi^{2n+1}v}$ equal to $\phi^{2n-1}$ in order to set up an induction and solve for $v$. After moving terms around and simplifying via identities involving $\phi$, we end up with $v = \frac{\phi^{2n-1}-\phi^{2n+1}}{1+\phi^{4n}} = -\frac{1}{L_{2n}}$ where $L_n$ is the $n^{th}$ Lucas number. As $arctan(-\frac{1}{x}) = -(2\;arctan(1)-arctan(x))$, this proves the claim for odd powers of $\phi$.

Q3: For even powers if we set up the analogous equation:

$arctan(\phi^{2n}) = arctan(\frac{\phi^{2n}+v}{1-\phi^{2n}v}) - arctan(v)$

However in this case, setting $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^{2n-2}$ and solving for $v$ gives $v=\frac{-1}{F_{2n-1}\sqrt{5}}$. This implies there cannot be a rational expression for $arctan(\phi^{2n})$ in terms of $arctan(\phi^{2n-2})$ and arctangents of integers for, if there were such an expression, the arctangent identity could then be applied to these terms to express $\frac{-1}{F_{2n-1}\sqrt{5}}$ as a rational number. Similar problems with $\sqrt{5}$ appearing in the expression for $v$ occur if we try setting the argument $\frac{\phi^{2n}+v}{1-\phi^{2n}v} = \phi^k$ for other powers $k$ such as $k=2n-1$, so if any $arctan(\phi^{2n})$ happen to be rational combinations of arctangents of integers, such expressions ought to be unrelated to one another unlike the situation for odd powers.

As a side note, this difference between the even and odd powers reminds me of the situation of values of the $\zeta$ function at integers where the $\zeta(2n)$ have simple closed form expressions while it is unknown if any $\zeta(2n+1)$ have a closed form expression. Perhaps someone with more familiarity of $\zeta$ values can comment on whether we might expect such similar behavior here.