Question

The sum of the squares two consecutive intergers is 145 . What are the intergers

Answer 1

Let the two consecutive integers be \( n \) and \( n + 1 \). The sum of the squares of these integers can be expressed as:

\[
n^2 + (n + 1)^2 = 145
\]

Now, expanding the equation:

\[
n^2 + (n^2 + 2n + 1) = 145
\]

Simplifying this gives:

\[
2n^2 + 2n + 1 = 145
\]

Subtracting 145 from both sides:

\[
2n^2 + 2n + 1 - 145 = 0
\]

This simplifies to:

\[
2n^2 + 2n - 144 = 0
\]

Dividing the entire equation by 2:

\[
n^2 + n - 72 = 0
\]

Next, we need to factor the quadratic equation. We are looking for two numbers that multiply to \(-72\) and add to \(1\). The factors are \(9\) and \(-8\):

\[
(n + 9)(n - 8) = 0
\]

Setting each factor equal to zero gives us:

\[
n + 9 = 0 \quad \Rightarrow \quad n = -9
\]
\[
n - 8 = 0 \quad \Rightarrow \quad n = 8
\]

Thus, the two pairs of consecutive integers are:

1. \( n = -9 \), \( n + 1 = -8 \)
2. \( n = 8 \), \( n + 1 = 9 \)

Therefore, the pairs of consecutive integers that satisfy the condition are:

\((-9, -8)\) and \((8, 9)\).