solution: We could find the answer by trial and error -- testing each candidate $t$ to see if $t\cdot (t+2)\equiv 1\pmod{23}$. However, here is another way:
  
  We can easily see that $4\cdot 6=24\equiv 1\pmod{23}$, so $4$ fulfills the main requirement that its inverse is $2$ more than it. Unfortunately, $4$ isn't odd. But we also have \begin{align*}
  (-4)\cdot (-6) &= 4\cdot 6 \\
  &\equiv 1\pmod{23},
  \end{align*} so $-4$ and $-6$ are each other's inverses $\pmod{23}$. Since $-4\equiv 19\pmod{23}$ and $-6\equiv 17\pmod{23}$, the answer $t=\boxed{17}$ satisfies the requirements of the problem.
  
  (We can even check that $17\cdot 19 = 323 = 14\cdot 23 + 1$.)
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Find the unique $\textbf{odd}$ integer $t$ such that $0<t<23$ and $t+2$ is the inverse of $t$ modulo $23$.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 17
extra_info.question: Find the unique $\textbf{odd}$ integer $t$ such that $0<t<23$ and $t+2$ is the inverse of $t$ modulo $23$.
extra_info.split: test