In maths class Albrecht had to compute $(a+2b-3)^2$ . His result was $a^2 +4b^2-9$ . ‘This is not correct’ said his teacher, ‘try substituting positive integers for $a$ and $b$.’ Albrecht did so, but his result proved to be correct. What numbers could he substitute?
a) Show a good substitution.
b) Give all the pairs that Albrecht could substitute and prove that there are no more.
To solve the problem, we need to verify the given expression and find all possible pairs \((a, b)\) such that Albrecht's result holds true.

1. **Expand the given expression:**
   \[
   (a + 2b - 3)^2
   \]
   Using the binomial expansion formula \((x + y)^2 = x^2 + 2xy + y^2\), we get:
   \[
   (a + 2b - 3)^2 = a^2 + 4b^2 + 4ab - 6a - 12b + 9
   \]

2. **Compare with Albrecht's result:**
   Albrecht's result is:
   \[
   a^2 + 4b^2 - 9
   \]
   For these two expressions to be equal, we must have:
   \[
   a^2 + 4b^2 + 4ab - 6a - 12b + 9 = a^2 + 4b^2 - 9
   \]

3. **Simplify the equation:**
   Subtract \(a^2 + 4b^2 - 9\) from both sides:
   \[
   a^2 + 4b^2 + 4ab - 6a - 12b + 9 - (a^2 + 4b^2 - 9) = 0
   \]
   This simplifies to:
   \[
   4ab - 6a - 12b + 18 = 0
   \]

4. **Rearrange the equation:**
   \[
   4ab - 6a - 12b + 18 = 0
   \]
   Divide the entire equation by 2:
   \[
   2ab - 3a - 6b + 9 = 0
   \]

5. **Use Simon's Favorite Factoring Trick:**
   Add and subtract 9 to factor the equation:
   \[
   2ab - 3a - 6b + 9 - 9 = 0
   \]
   \[
   2ab - 3a - 6b = -9
   \]
   Factor by grouping:
   \[
   (2a - 3)(2b - 3) = 27
   \]

6. **Find integer solutions:**
   We need to find pairs \((a, b)\) such that \((2a - 3)(2b - 3) = 27\). The factor pairs of 27 are:
   \[
   (1, 27), (3, 9), (9, 3), (27, 1)
   \]

   Solving for \(a\) and \(b\) in each case:
   - For \((2a - 3) = 1\) and \((2b - 3) = 27\):
     \[
     2a - 3 = 1 \implies 2a = 4 \implies a = 2
     \]
     \[
     2b - 3 = 27 \implies 2b = 30 \implies b = 15
     \]
     So, \((a, b) = (2, 15)\).

   - For \((2a - 3) = 3\) and \((2b - 3) = 9\):
     \[
     2a - 3 = 3 \implies 2a = 6 \implies a = 3
     \]
     \[
     2b - 3 = 9 \implies 2b = 12 \implies b = 6
     \]
     So, \((a, b) = (3, 6)\).

   - For \((2a - 3) = 9\) and \((2b - 3) = 3\):
     \[
     2a - 3 = 9 \implies 2a = 12 \implies a = 6
     \]
     \[
     2b - 3 = 3 \implies 2b = 6 \implies b = 3
     \]
     So, \((a, b) = (6, 3)\).

   - For \((2a - 3) = 27\) and \((2b - 3) = 1\):
     \[
     2a - 3 = 27 \implies 2a = 30 \implies a = 15
     \]
     \[
     2b - 3 = 1 \implies 2b = 4 \implies b = 2
     \]
     So, \((a, b) = (15, 2)\).

Conclusion:
The pairs \((a, b)\) that Albrecht could substitute are \((2, 15), (3, 6), (6, 3), (15, 2)\).

The final answer is \( \boxed{ (2, 15), (3, 6), (6, 3), (15, 2) } \).