What is the sum of all integer values of n for which n 2+ 19n + 92 is a perfect square?
-19
Understanding the Problem: Perfect Square Condition
The question asks for the sum of all integer values of \(n\) such that the expression \(n^2 + 19n + 92\) results in a perfect square. A perfect square is an integer that is the square of another integer. For example, 4, 9, 16 are perfect squares (\(2^2, 3^2, 4^2\)).
Let the given expression be equal to a perfect square, say \(k^2\), where \(k\) is an integer.
So, we have the equation:
\(n^2 + 19n + 92 = k^2\)
We need to find all integer values of \(n\) that satisfy this equation and then calculate their sum.
Solving the Equation Using Algebraic Manipulation
To solve this equation, we can use the method of completing the square for the variable \(n\). The expression \(n^2 + 19n\) is part of a perfect square trinomial \((n+a)^2 = n^2 + 2an + a^2\). Comparing \(n^2 + 19n\) with \(n^2 + 2an\), we have \(2a = 19\), so \(a = \frac{19}{2}\). The term needed to complete the square is \(a^2 = \left(\frac{19}{2}\right)^2 = \frac{361}{4}\).
Add and subtract \(\frac{361}{4}\) to the expression:
\(n^2 + 19n + \frac{361}{4} - \frac{361}{4} + 92 = k^2\)
\(\left(n + \frac{19}{2}\right)^2 - \frac{361}{4} + \frac{92 \times 4}{4} = k^2\)
\(\left(n + \frac{19}{2}\right)^2 - \frac{361}{4} + \frac{368}{4} = k^2\)
\(\left(n + \frac{19}{2}\right)^2 + \frac{7}{4} = k^2\)
To eliminate the fractions, multiply the entire equation by 4:
\(4 \left(n + \frac{19}{2}\right)^2 + 4 \times \frac{7}{4} = 4k^2\)
Recognize that \(4 \left(n + \frac{19}{2}\right)^2 = \left(2 \left(n + \frac{19}{2}\right)\right)^2 = (2n + 19)^2\).
So the equation becomes:
\((2n + 19)^2 + 7 = (2k)^2\)
Rearrange the terms to form a difference of squares:
\((2k)^2 - (2n + 19)^2 = 7\)
Let \(A = 2k\) and \(B = 2n + 19\). The equation is \(A^2 - B^2 = 7\).
Using the difference of squares formula, \(A^2 - B^2 = (A-B)(A+B)\), we get:
\((2k - (2n + 19))(2k + (2n + 19)) = 7\)
\((2k - 2n - 19)(2k + 2n + 19) = 7\)
Finding Integer Solutions using Factor Pairs
Since \(n\) is an integer, \(2n + 19\) is an integer. Since \(k\) is an integer (because \(k^2\) is a perfect square), \(2k\) is an integer. Therefore, the two factors \((2k - 2n - 19)\) and \((2k + 2n + 19)\) must be integers.
We need to find pairs of integers whose product is 7. The integer factor pairs of 7 are:
- (1, 7)
- (7, 1)
- (-1, -7)
- (-7, -1)
Let the two factors be \(X = 2k - 2n - 19\) and \(Y = 2k + 2n + 19\), such that \(XY = 7\).
We can solve for \(2n + 19\) and \(2k\) by adding and subtracting these two equations:
- \(Y + X = (2k + 2n + 19) + (2k - 2n - 19) = 4k\)
- \(Y - X = (2k + 2n + 19) - (2k - 2n - 19) = 4n + 38\)
This implies that \(X\) and \(Y\) must have the same parity. Since their product 7 is odd, both \(X\) and \(Y\) must be odd. All the factor pairs listed above consist of odd numbers, so all cases are possible.
Case 1: \(2k - 2n - 19 = 1\) and \(2k + 2n + 19 = 7\)
Adding the two equations:
\((2k - 2n - 19) + (2k + 2n + 19) = 1 + 7\)
\(4k = 8 \implies k = 2\)
Subtracting the first equation from the second:
\((2k + 2n + 19) - (2k - 2n - 19) = 7 - 1\)
\(4n + 38 = 6\)
\(4n = 6 - 38\)
\(4n = -32\)
\(n = -8\)
Let's check this value of n: \((-8)^2 + 19(-8) + 92 = 64 - 152 + 92 = 156 - 152 = 4\). \(4 = 2^2\), which is a perfect square. So \(n=-8\) is a valid integer value.
Case 2: \(2k - 2n - 19 = 7\) and \(2k + 2n + 19 = 1\)
Adding the two equations:
\(4k = 7 + 1 = 8 \implies k = 2\)
Subtracting the first equation from the second:
\(4n + 38 = 1 - 7 = -6\)
\(4n = -6 - 38\)
\(4n = -44\)
\(n = -11\)
Let's check this value of n: \((-11)^2 + 19(-11) + 92 = 121 - 209 + 92 = 213 - 209 = 4\). \(4 = 2^2\), which is a perfect square. So \(n=-11\) is a valid integer value.
Case 3: \(2k - 2n - 19 = -1\) and \(2k + 2n + 19 = -7\)
Adding the two equations:
\(4k = -1 + (-7) = -8 \implies k = -2\)
Subtracting the first equation from the second:
\(4n + 38 = -7 - (-1) = -6\)
\(4n = -6 - 38\)
\(4n = -44\)
\(n = -11\)
This gives the same integer value \(n=-11\). Check: \((-11)^2 + 19(-11) + 92 = 4 = (-2)^2\), a perfect square.
Case 4: \(2k - 2n - 19 = -7\) and \(2k + 2n + 19 = -1\)
Adding the two equations:
\(4k = -7 + (-1) = -8 \implies k = -2\)
Subtracting the first equation from the second:
\(4n + 38 = -1 - (-7) = 6\)
\(4n = 6 - 38\)
\(4n = -32\)
\(n = -8\)
This gives the same integer value \(n=-8\). Check: \((-8)^2 + 19(-8) + 92 = 4 = (-2)^2\), a perfect square.
The integer values of \(n\) for which \(n^2 + 19n + 92\) is a perfect square are -8 and -11.
Calculating the Sum of Integer Values of n
The question asks for the sum of all such integer values of \(n\). The integer values we found are -8 and -11.
Sum = \(-8 + (-11)\)
Sum = \(-8 - 11\)
Sum = \(-19\)
Summary of Results
| Factor Pair (X, Y) | Equation System | Resulting k | Resulting n | Is n an integer? | Is n^2 + 19n + 92 a perfect square? |
|---|---|---|---|---|---|
| (1, 7) | \(2k - 2n - 19 = 1\), \(2k + 2n + 19 = 7\) | 2 | -8 | Yes | \(4 = 2^2\) (Yes) |
| (7, 1) | \(2k - 2n - 19 = 7\), \(2k + 2n + 19 = 1\) | 2 | -11 | Yes | \(4 = 2^2\) (Yes) |
| (-1, -7) | \(2k - 2n - 19 = -1\), \(2k + 2n + 19 = -7\) | -2 | -11 | Yes | \(4 = (-2)^2\) (Yes) |
| (-7, -1) | \(2k - 2n - 19 = -7\), \(2k + 2n + 19 = -1\) | -2 | -8 | Yes | \(4 = (-2)^2\) (Yes) |
The integer values of \(n\) found are -8 and -11. The sum of these values is -8 + (-11) = -19.
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Perfect Square | An integer that is the square of another integer (\(m^2\)). | The condition \(n^2 + 19n + 92 = k^2\). |
| Completing the Square | An algebraic technique to rewrite a quadratic expression in the form \((x+a)^2 + b\). | Used to transform \(n^2 + 19n + 92\) into a form involving \((n + 19/2)^2\). |
| Difference of Squares | The algebraic identity \(a^2 - b^2 = (a-b)(a+b)\). | Used to factorize \((2k)^2 - (2n+19)^2\). |
| Integer Solutions | Values of variables that are integers. | We are looking for integer values of \(n\) and \(k\). |
| Factor Pairs | A pair of integers whose product equals a specific integer. | Used to find possible values for \((2k - 2n - 19)\) and \((2k + 2n + 19)\). |
Additional Information: Diophantine Equations
The problem we solved is a type of Diophantine equation, which is a polynomial equation where the variables are restricted to be integers. The equation \(n^2 + 19n + 92 = k^2\) is an example. By transforming it into \((2k)^2 - (2n + 19)^2 = 7\), we converted it into a simpler form \(A^2 - B^2 = C\), where we seek integer solutions for \(A\) and \(B\). This specific form can often be solved by factoring the difference of squares and considering the factors of \(C\). The requirement that \(A\) and \(B\) must satisfy certain parity or other conditions based on their definitions (\(A=2k, B=2n+19\)) helps narrow down the possibilities from the factor pairs.
In our case, \(A = 2k\) means A is even, and \(B = 2n + 19\) means B is odd (since 2n is even, 2n+19 is odd). However, the variables introduced \(X = 2k - 2n - 19\) and \(Y = 2k + 2n + 19\) must both be odd because their sum \(X+Y=4k\) and difference \(Y-X=4n+38\) are even. Our analysis based on \(X\) and \(Y\) being odd was correct and led to the valid integer solutions for \(n\).
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