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For how many integer values of \(n\) is \(n^2 + 6n - 8\) the square of an integer?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
23 Jan 2024, 02:24
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Official Solution:
For how many integer values of \(n\) is \(n^2 + 6n - 8\) the square of an integer?
A. 0
B. 1
C. 2
D. 3
E. 4
We need to find how many integer values of \(n\) exist such that \(n^2 + 6n - 8=m^2\), for some integer \(m\).
Complete the square for the expression on the left-hand side:
\(n^2 + 6n + 9 - 17=m^2\)
\((n + 3)^2 - 17=m^2\)
Re-arrange:
\((n + 3)^2 - m^2 = 17\)
Apply the difference of squares formula:
\((n + 3 - m)(n + 3 + m) = 17\)
So, we find that the product of two integers, \((n + 3 - m)\) and \((n + 3 + m)\), equals the prime number 17. 17 can be expressed as the product of two integers in only two ways: \(17 = 1*17\) and \(17 = (-1)(-17)\).
\((n + 3 - m)=1\) and \((n + 3 + m) =17\) or vice versa gives \(n=6\).
\((n + 3 - m)=-1\) and \((n + 3 + m) =-17\) or vice versa gives \(n=-12\).
Therefore, there are two such values of \(n\) for which \(n^2 + 6n - 8\) is the square of an integer.
Answer: C
For how many integer values of \(n\) is \(n^2 + 6n - 8\) the square of an integer?
A. 0
B. 1
C. 2
D. 3
E. 4
We need to find how many integer values of \(n\) exist such that \(n^2 + 6n - 8=m^2\), for some integer \(m\).
Complete the square for the expression on the left-hand side:
\(n^2 + 6n + 9 - 17=m^2\)
\((n + 3)^2 - 17=m^2\)
Re-arrange:
\((n + 3)^2 - m^2 = 17\)
Apply the difference of squares formula:
\((n + 3 - m)(n + 3 + m) = 17\)
So, we find that the product of two integers, \((n + 3 - m)\) and \((n + 3 + m)\), equals the prime number 17. 17 can be expressed as the product of two integers in only two ways: \(17 = 1*17\) and \(17 = (-1)(-17)\).
\((n + 3 - m)=1\) and \((n + 3 + m) =17\) or vice versa gives \(n=6\).
\((n + 3 - m)=-1\) and \((n + 3 + m) =-17\) or vice versa gives \(n=-12\).
Therefore, there are two such values of \(n\) for which \(n^2 + 6n - 8\) is the square of an integer.
Answer: C
