a and b are positive integers that `a^(2) + 2b = b^(2) + 2a +5`. The value of b is…………….

Answer

To solve the equation \( a^2 + 2b = b^2 + 2a + 5 \) for positive integers \( a \) and \( b \), we can follow these steps:

### Step 1: Rearrange the equation
Start by rearranging the equation to group similar terms:
\[ a^2 - b^2 + 2b - 2a - 5 = 0 \]

### Step 2: Factor the equation
Notice that \( a^2 - b^2 \) can be factored using the difference of squares:
\[ (a - b)(a + b) + 2b - 2a - 5 = 0 \]

### Step 3: Complete the square
To make the equation easier to handle, we can complete the square for both \( a \) and \( b \):
\[ (a - 1)^2 - (b - 1)^2 = 5 \]
This can be rewritten as:
\[ (a - 1 - (b - 1))(a - 1 + (b - 1)) = 5 \]
or
\[ (a - b)(a + b - 2) = 5 \]

### Step 4: Analyze the factors of 5
Since \( a \) and \( b \) are positive integers, we can consider the possible pairs of factors of 5:
1. \( (1, 5) \)
2. \( (5, 1) \)

### Step 5: Case 1: \( a - b = 1 \) and \( a + b - 2 = 5 \)
From \( a - b = 1 \):
\[ a = b + 1 \]
Substituting into the second equation:
\[ (b + 1) + b - 2 = 5 \Rightarrow 2b - 1 = 5 \Rightarrow 2b = 6 \Rightarrow b = 3 \]
Then,
\[ a = b + 1 = 3 + 1 = 4 \]

### Step 6: Case 2: \( a - b = 5 \) and \( a + b - 2 = 1 \)
From \( a - b = 5 \):
\[ a = b + 5 \]
Substituting into the second equation:
\[ (b + 5) + b - 2 = 1 \Rightarrow 2b + 3 = 1 \Rightarrow 2b = -2 \Rightarrow b = -1 \]
This case is not valid since \( b \) must be a positive integer.

### Conclusion
The only valid solution from the cases is:
\[ b = 3 \]
Thus, the value of \( b \) is \( \boxed{3} \).