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The value of the equation √(n² - 19n - 99) is an integer and n is also an integer then find the value of n. In other words, (n² - 19n - 99) is a perfect square and n belongs to an integer. Find the value of n.
Solution 1:
Let,
\[\sqrt{n^2-19n-99}=x\]
\[\Rightarrow n^2-19n-99=x^2\]
\[\Rightarrow n^2-19n-(99+x^2)=0\]
It is a quadratic equation,
\[\Rightarrow n=\frac{19\pm\sqrt{361+4(99+x^2)}}{2}\]
\[\Rightarrow n=\frac{19\pm\sqrt{757+4x^2}}{2}\]
n is an integer (757+4x^2 must be perfect square),
\[\Rightarrow 757+4x^2=k^2\]
\[\Rightarrow 757=k^2-4x^2\]
\[\Rightarrow 1\times757=(k-2x)(k+2x)\]
757 don't have any other factors,
Let,
k - 2x = 1
k + 2x = 757
⇒ k = 379, x = 189
\[\Rightarrow n=\frac{19\pm k}{2}=\frac{19\pm379}{2}\]
⇒ n = 199 or n = -180
Solution 2:
Let,
\[\sqrt{n^2-19n-99}=k\]
\[\Rightarrow n^2-19n-99=k^2\]
\[\Rightarrow n^2-2\times\frac{19}{2}n+\left(\frac{19}{2}\right)^2-99-\left(\frac{19}{2}\right)^2=k^2\]
\[\Rightarrow \left(n-\frac{19}{2}\right)^2-\frac{757}{4}=k^2\]
\[\Rightarrow \left(2n-19\right)^2-757=(2k)^2\]
\[\Rightarrow \left(2n-19\right)^2-(2k)^2=757\]
\[\Rightarrow (2n-19-2k)(2n-19+2k)=1\times757\]
757 don't have any other factors,\
Let,
2n - 19 - 2k = 1
2n - 19 + 2k = 757
⇒ n = 199
Now,
2n - 19 - 2k = -1
2n - 19 + 2k = -757
⇒ n = -180
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