<Question>
Find the unique $\textbf{odd}$ integer $t$ such that $0<t<23$ and $t+2$ is the inverse of $t$ modulo $23$.
</Question>

<think>
<Plan_1>
Okay, let's try to solve this problem. So, we need to find an odd integer t between 1 and 22 (since 0 < t < 23) such that t + 2 is the inverse of t modulo 23. Hmm, inverses modulo 23... right, the inverse of a number a modulo m is a number b such that a * b ≡ 1 mod m. So in this case, t * (t + 2) ≡ 1 mod 23. That gives us an equation to solve: t(t + 2) ≡ 1 mod 23. Let me write that down properly.
So, the equation is t(t + 2) ≡ 1 mod 23.

Expanding that, we get t² + 2t ≡ 1 mod 23.

Subtracting 1 from both sides, we have t² + 2t - 1 ≡ 0 mod 23.
So, this is a quadratic congruence equation: t² + 2t - 1 ≡ 0 mod 23.

To solve quadratic congruences, maybe I can complete the square. Let's try that.

<prompt_1.1> Solve the quadratic congruence equation by completing the square and find the solutions for t. </prompt_1.1>
-----

But let me check if there are other solutions. Wait, quadratic equations modulo primes can have at most two solutions.
Since 23 is prime, and we found two solutions, t=5 and t=21, but only t=5 is odd. So that's the unique solution.

But just to be thorough, let me check all odd numbers between 1 and 22 to make sure. Although that might take longer, but perhaps I can check a few to confirm.

<prompt_1.2> Check all odd numbers between 1 and 22 to confirm the solution. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The equation is t² + 2t - 1 ≡ 0 mod 23. Let's complete the square. 

t² + 2t ≡ 1 mod 23.

Adding 1 to both sides to complete the square: 

t² + 2t + 1 ≡ 1 + 1 mod 23.

So, (t + 1)² ≡ 2 mod 23.

Therefore, we need to find t such that (t + 1)² ≡ 2 mod 23. That means t + 1 must be a square root of 2 modulo 23.
</execution_1.1>

<execution_1.2>
For example, t=1: inverse of 1 mod23 is 1, but t+2=3, which is not 1. So no. t=3: inverse of 3 mod23 is 8 because 3*8=24≡1 mod23. t+2=5, which is not 8. t=5: inverse is 14, and t+2=7. Wait, 5+2=7, but inverse is 14. Wait, that's not matching. Wait, hold on. Wait, inverse of t is t+2. So for t=5, inverse is 7. But 5*7=35≡35-23=12≡12 mod23. That's not 1. Wait, that contradicts earlier result. Wait, did I make a mistake here?

Wait, let me check again. If t=5, then t+2=7. So according to the problem, 7 should be the inverse of 5 modulo23. So 5*7 should be ≡1 mod23. But 5*7=35, and 35 mod23 is 12, not 1. That's not correct. Wait, that can't be. So maybe I messed up somewhere.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, we need to find the square roots of 2 modulo 23. Let me recall that 23 is a prime, so the equation x² ≡ 2 mod 23 may have two solutions. Let's try to find them.

<prompt_2.1> Find the square roots of 2 modulo 23 and determine the corresponding values of t. </prompt_2.1>
-----

Based on execution_1.2:
Wait, that's confusing. Because earlier when we solved the equation, we found t=5 and t=21. But when I checked t=5, 5*7=35≡12 mod23, which is not 1. So that's a problem.

<prompt_2.2> Verify the solution t=5 by checking if 5*7 ≡ 1 mod 23. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's test some numbers. Let's see:

1² = 1 mod 23

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 2 mod 23. Oh, so 5 is a square root of 2 modulo 23. Then, the other square root would be -5 mod 23, which is 18, since 23 - 5 = 18. Let's check 18²: 18² = 324. 324 divided by 23: 23*14 = 322, so 324 - 322 = 2. So 18² ≡ 2 mod 23. Correct. So the square roots of 2 mod 23 are 5 and 18.

Therefore, (t + 1) ≡ 5 mod 23 or (t + 1) ≡ 18 mod 23.

Solving for t:

Case 1: t + 1 ≡ 5 mod 23 ⇒ t ≡ 5 - 1 ≡ 4 mod 23.

Case 2: t + 1 ≡ 18 mod 23 ⇒ t ≡ 18 - 1 ≡ 17 mod 23.

But the problem states that t must be an odd integer between 1 and 22. Let's check the solutions:

t ≡ 4 mod 23: 4 is even, so not acceptable.

t ≡ 17 mod 23: 17 is odd and 17 is between 1 and 22. So t = 17.
</execution_2.1>

<execution_2.2>
So where did I go wrong here? Let me check my steps again. 

Starting from the beginning: t is odd, 0 < t <23. t + 2 is inverse of t modulo23. So t*(t + 2) ≡1 mod23. Expanding: t² + 2t ≡1 mod23. Then completing the square: t² + 2t +1 ≡2 mod23, so (t +1)^2 ≡2 mod23. Therefore, t +1 ≡ sqrt(2) or -sqrt(2) mod23. So t ≡ -1 + sqrt(2) or -1 - sqrt(2) mod23. 

But if sqrt(2) mod23 is 13 and 10, as we found earlier, then t would be -1 +13=12 or -1 +10=9. But 12 and 9 are even, which contradicts the requirement that t is odd. Therefore, there is no solution? But the problem states there is a unique odd integer t. So where is the mistake here?
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but let's check if t=17 satisfies the original condition. Let's compute t + 2 = 19. Then check if 19 is the inverse of 17 modulo23.

<prompt_3.1> Verify if t = 17 satisfies the original condition by checking if 19 is the inverse of 17 modulo 23. </prompt_3.1>
-----

Based on execution_2.2:
Wait, maybe I miscalculated the square roots of 2 mod23. Let me check again.

<prompt_3.2> Recalculate the square roots of 2 modulo 23 to verify the previous steps. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Compute 17*19: 17*20=340, subtract 17 to get 340 -17=323. Now divide 323 by 23: 23*14=322, so 323 -322=1. Therefore, 17*19 ≡1 mod23. Yes, that works! So t=17 is a solution. But wait, the problem says "the unique odd integer t". But 17 is odd, that's okay. Wait, but wait, 17 is between 1 and 22, and it's odd. So maybe the answer is 17?
</execution_3.1>

<execution_3.2>
Let's compute 5²=25≡2 mod23, yes. Then 23 -5=18, so 18²=(23 -5)²=25≡2 mod23. So 18 is the other root. So that's correct. Then t +1 ≡5 or 18. So t=4 or 17. 4 is even, 17 is odd. So t=17. But then t=17, inverse is 19. Let's check 17*19. 17*20=340, minus 17 is 323. 323 divided by23: 23*14=322, so 323-322=1. So 17*19=323≡1 mod23. Yes, that works. So t=17 is a solution. 

But the problem says "the unique odd integer t". Wait, but earlier when I thought t=5, that was a mistake because I thought t=5 was a solution, but actually t=17 is the correct one. So where did I get t=5 from? Maybe I confused the inverse. Let me check t=5. If t=5, then t +2=7. Then inverse of 5 is 7. So 5*7=35≡35-23=12≡12 mod23≠1. So that's not valid. So t=5 is not a solution. So the correct solution is t=17. 

But then why did I earlier think t=5? Maybe because 5 is a square root of 2, but in the equation (t +1)^2≡2, so t +1≡5 or 18, leading to t=4 or 17. So t=17 is the correct one. 

So perhaps the answer is t=17. But the problem states "the unique odd integer t". So maybe 17 is the answer. But then why did I think of t=5 earlier? Maybe confusion. 

Wait, but let's check t=17: inverse is 19, and t +2=19. So yes, 19 is the inverse. So 17 is the solution. 

But the problem says "unique odd integer t" such that 0 < t <23. So t=17 is the answer.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, the problem says "the unique odd integer t". But 17 is odd, that's okay. Wait, but wait, 17 is between 1 and 22, and it's odd. So maybe the answer is 17?

Wait, but let's check if there are other solutions. Wait, the quadratic equation modulo 23 can have at most two solutions. We found t=4 and t=17. But t=4 is even, so we discard it. So t=17 is the only solution. Therefore, the answer is 17.
-----

Based on execution_3.2:
But let me check once more for t=17: 17 is odd, 0 <17<23.

<prompt_4.1> Perform a final verification of the solution t = 17 and conclude the answer. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
t +2=19. 17*19=323. 323 divided by23: 23*14=322, 323-322=1. So yes, 17*19≡1 mod23. Correct. So t=17 is the solution. 

But the problem says "the unique odd integer t". So 17 is the answer. 

Wait, but in the initial problem statement, when I first saw t=5, I thought maybe that's the answer, but then realized that 5*7=35≡12 mod23. So that's wrong. So the correct answer is 17. 

Therefore, after going through the steps, solving the quadratic congruence, checking the solutions, verifying the inverse, the answer is 17. 

But just to ensure no step was missed, let's recap:

1. Given t is odd, 0 < t <23, and t +2 is inverse of t mod23.
2. So t(t +2)≡1 mod23.
3. Which leads to t² +2t -1≡0 mod23.
4. Completing the square: (t +1)^2 ≡2 mod23.
5. Find square roots of 2 mod23: 5 and 18.
6. Thus, t +1≡5 ⇒ t=4 (even, discard) or t +1≡18 ⇒ t=17 (odd, accept).
7. Verify t=17: 17*19=323≡1 mod23. Correct.
8. Hence, t=17 is the unique solution. 

Therefore, the answer is 17.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
**Final Answer**
The unique odd integer \( t \) is \boxed{17}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the unique odd integer $t$ such that $0 < t < 23$ and $t + 2$ is the inverse of $t$ modulo $23$ is $t = \boxed{17}$.
</Final_answer>
</think>