Determine all values of $n^2 + 19n + 99$ is a perfect square. I tried setting some square $b^2$ equal to the following, and then factoring as a Diophantine equation with $2$ variables... Didn't work.
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Hint: $n^2 + 19 n + 99$ is a perfect square if and only if $4(n^2 + 19n + 99)$ is one. – Daniel Fischer Nov 06 '14 at 22:20
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Why times by 4? – user164403 Nov 06 '14 at 22:22
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Because it makes it easier to handle when setting it $= b^2$. – Daniel Fischer Nov 06 '14 at 22:23
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Completing the square becomes more pleasant. Fractions are hard. – André Nicolas Nov 06 '14 at 22:27
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$$ \begin{array}{c} n^2 + 19n+99 = r^2 \\ (4n^2+76n) = 4r^2 - 396 \\ (2n+19)^2 -361 = 4r^2 - 396 \\ r^2 - (2n+19)^2 = 35 \\ (r+ (2n+19)) (r-(2n+19)) = 35 \end{array} $$ Then if: $$ \begin{array}{ccc} (r+ (2n+19)) =35, & (r-(2n+19)) = 1 & \longrightarrow n = -1 \\ (r+ (2n+19)) =7, & (r-(2n+19)) = 5 & \longrightarrow n = -9 \\ (r+ (2n+19)) =5, & (r-(2n+19)) = 7 & \longrightarrow n = -10 \\ (r+ (2n+19)) =1, & (r-(2n+19)) = 35 & \longrightarrow n = -18 \\ (r+ (2n+19)) =-35, & (r-(2n+19)) = -1 & \longrightarrow n = -18 \\ (r+ (2n+19)) =-7, & (r-(2n+19)) = -5 & \longrightarrow n = -10 \\ (r+ (2n+19)) =-5, & (r-(2n+19)) = -7 & \longrightarrow n = -9 \\ (r+ (2n+19)) =-1, & (r-(2n+19)) = -5 & \longrightarrow n = -1 \\ \end{array} $$ So only those four integer solutions work. $n$ must be negative, by the way.
Mark Fischler
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