Let's denote the two consecutive integers as \( n \) and \( n + 1 \). The equation based on the problem's statement is:
\[ n^2 + (n + 1)^2 = 45 \]
Now, expand the equation:
\[ n^2 + (n^2 + 2n + 1) = 45 \]
Combine like terms:
\[ 2n^2 + 2n + 1 = 45 \]
Subtract 45 from both sides:
\[ 2n^2 + 2n + 1 - 45 = 0 \]
This simplifies to:
\[ 2n^2 + 2n - 44 = 0 \]
Now, divide the entire equation by 2:
\[ n^2 + n - 22 = 0 \]
Next, we can factor this quadratic equation (or use the quadratic formula). We need two numbers that multiply to \( -22 \) and add to \( 1 \). The numbers \( 11 \) and \( -2 \) fit, so we can factor it as:
\[ (n + 11)(n - 2) = 0 \]
Setting each factor equal to zero gives us the possible solutions for \( n \):
\[ n + 11 = 0 \quad \Rightarrow \quad n = -11 \]
\[ n - 2 = 0 \quad \Rightarrow \quad n = 2 \]
Thus, the pairs of consecutive integers are:
- For \( n = -11 \): The integers are \( -11 \) and \( -10 \).
- For \( n = 2 \): The integers are \( 2 \) and \( 3 \).
Therefore, the consecutive integers that satisfy the equation are \( (-11, -10) \) and \( (2, 3) \).
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