Okay, so I need to find an odd integer t between 1 and 22 such that when I add 2 to it, the result is the inverse of t modulo 23. Hmm, let me think about how to approach this.

First, let's recall what it means for two numbers to be inverses modulo 23. If t and t+2 are inverses modulo 23, then their product should be congruent to 1 modulo 23. So mathematically, that means t*(t + 2) ≡ 1 mod 23. Let me write that down:

t(t + 2) ≡ 1 mod 23.

Expanding the left side, that gives t² + 2t ≡ 1 mod 23. Rearranging terms, we get a quadratic congruence:

t² + 2t - 1 ≡ 0 mod 23.

So the problem reduces to solving this quadratic equation modulo 23. Since we're dealing with a quadratic equation modulo a prime (23 is prime), there can be at most two solutions. However, the problem specifies that t must be an odd integer between 1 and 22. So we need to check which of the solutions (if any) are odd and within that range.

Let me recall how to solve quadratic congruences modulo primes. The general approach is to complete the square or use the quadratic formula for modular arithmetic. Let me try completing the square first.

Starting with t² + 2t - 1 ≡ 0 mod 23. Let's move the constant term to the other side:

t² + 2t ≡ 1 mod 23.

To complete the square, we take the coefficient of t, which is 2, divide by 2 to get 1, then square it to get 1. So we add 1 to both sides:

t² + 2t + 1 ≡ 1 + 1 mod 23.

This simplifies to:

(t + 1)² ≡ 2 mod 23.

So now we have (t + 1)² ≡ 2 mod 23. Therefore, t + 1 must be a square root of 2 modulo 23. So we need to find all x such that x² ≡ 2 mod 23, and then t = x - 1.

But first, does 2 have square roots modulo 23? Let me check if 2 is a quadratic residue modulo 23. The Legendre symbol (2|23) can help here. The Legendre symbol (a|p) is 1 if a is a quadratic residue modulo an odd prime p, -1 if it's a non-residue, and 0 if a is divisible by p.

Using the formula for the Legendre symbol (2|p) when p is an odd prime:

(2|p) = 1 if p ≡ ±1 mod 8,
(2|p) = -1 if p ≡ ±3 mod 8.

23 divided by 8 is 2 with a remainder of 7, so 23 ≡ 7 mod 8, which is equivalent to -1 mod 8. Therefore, (2|23) = -1, meaning 2 is a quadratic non-residue modulo 23. Wait, that contradicts my initial thought. Wait, 23 mod 8 is 7, which is 7 ≡ -1 mod 8, so (2|23) = -1. Therefore, 2 is a non-residue modulo 23. That means that the equation x² ≡ 2 mod 23 has no solutions. Hmm, that can't be right. Wait, but 2 is a quadratic residue modulo some primes. Let me verify.

Wait, maybe I made a mistake in the formula. Let me check again. The Legendre symbol (2|p) is 1 if p ≡ 1 or 7 mod 8, and -1 if p ≡ 3 or 5 mod 8. Since 23 ≡ 7 mod 8, then (2|23) = 1. Wait, that's different from what I thought before. Wait, maybe I confused the formula.

Wait, let's recall that (2|p) = 1 if p ≡ 1 or 7 mod 8, and -1 otherwise. So since 23 ≡ 7 mod 8, then (2|23) = 1. Therefore, 2 is a quadratic residue modulo 23. Therefore, there are solutions.

Wait, maybe I messed up the formula. Let me check with a small example. For p=7, which is 7 mod 8, (2|7) should be 1. Indeed, 3²=9≡2 mod7. So yes, 2 is a quadratic residue modulo 7. Similarly, 23 is 7 mod 8, so 2 should be a quadratic residue modulo 23. Therefore, there are solutions.

Therefore, x² ≡ 2 mod 23 has solutions. So let's find them. We need to find integers x such that x² ≡ 2 mod23. Let's try to compute x.

One way is to test small numbers. Let's compute squares modulo 23:

1²=1

2²=4

3²=9

4²=16

5²=25≡2 mod23. Oh, 5²=25≡2 mod23. So x=5 is a solution. Then the other solution is -5 mod23, which is 18, since 5+18=23. So x=5 and x=18 are solutions.

Therefore, (t + 1) ≡5 mod23 or (t + 1)≡18 mod23. Therefore, t ≡5 -1=4 mod23 or t≡18 -1=17 mod23.

So the solutions for t are t≡4 mod23 and t≡17 mod23. Now, since we're looking for t such that 0 < t < 23, the possible values are t=4 and t=17. However, the problem states that t must be odd. So let's check which of these are odd.

t=4 is even, so that's out. t=17 is odd, so that's the candidate. Therefore, t=17 is the solution. Let me verify this.

Compute t+2=17+2=19. Now check if 19 is the inverse of 17 modulo23.

Compute 17*19 mod23. 17*19=323. Divide 323 by23: 23*14=322, so 323-322=1. Therefore, 17*19≡1 mod23. Correct. So indeed, 19 is the inverse of17 modulo23. Therefore, t=17 is the solution.

Wait, but let me check if there are other solutions. Since the quadratic congruence had two solutions, t=4 and t=17. But only t=17 is odd. So that's the unique solution. Therefore, the answer is 17.

But wait, just to make sure I didn't make any mistakes in the quadratic formula step. Let me go through the steps again.

Starting with t(t + 2) ≡1 mod23 => t² +2t -1 ≡0 mod23. Completing the square: t² +2t +1 ≡2 mod23 => (t +1)^2 ≡2 mod23. So (t +1)^2 ≡2 mod23. Then, as 2 is a quadratic residue modulo23, the solutions are t+1 ≡5 or 18 mod23, leading to t=4 or17. Only 17 is odd. All steps seem correct. So yes, t=17 is the solution.

Alternatively, maybe there's another way to approach this problem. For example, since t is odd, let's list all the odd numbers between1 and22 and check which one satisfies t+2 ≡ inverse of t mod23. There are 11 odd numbers:1,3,5,7,9,11,13,15,17,19,21. For each, compute t+2 and check if (t+2)*t ≡1 mod23.

Let's do that:

1. t=1: t+2=3. 1*3=3≡3≠1 mod23. Not inverse.
2. t=3: t+2=5. 3*5=15≡15≠1 mod23.
3. t=5: t+2=7. 5*7=35≡35-23=12≡12≠1 mod23.
4. t=7: t+2=9. 7*9=63≡63-2*23=63-46=17≡17≠1.
5. t=9: t+2=11. 9*11=99≡99-4*23=99-92=7≡7≠1.
6. t=11: t+2=13. 11*13=143≡143-6*23=143-138=5≡5≠1.
7. t=13: t+2=15. 13*15=195≡195-8*23=195-184=11≡11≠1.
8. t=15: t+2=17. 15*17=255≡255-11*23=255-253=2≡2≠1.
9. t=17: t+2=19. 17*19=323≡323-14*23=323-322=1≡1. Bingo.
10. t=19: t+2=21. 19*21=399≡399-17*23=399-391=8≡8≠1.
11. t=21: t+2=23≡0 mod23. 21*0=0≡0≠1.

So indeed, only t=17 works. Therefore, the unique odd integer t between1 and22 is17. So the answer is17. That matches the previous result. Therefore, I can be confident that the answer is correct.

Another way to check: Since t and t+2 are inverses, then t*(t+2)≡1 mod23. Let's substitute t=17:17*19=323. Divided by23:23*14=322, so 323-322=1. So 17*19=1 mod23. Correct.

Therefore, all methods confirm that t=17 is the solution. So I think that's solid.
Therefore, the unique odd integer $t$ satisfying the given conditions is $\boxed{17}$.
</think>