# √(n^2-9n-1) 還是正整數？

## 解答

$\sqrt{n^2-9n-1}=t$，其中 t 為正整數

$n^2-9n-1=t^2$

$n^2-9n+\left( \frac{9}{2} \right)^2-1=t^2+\left( \frac{9}{2} \right)^2$

$\left( n-\frac{9}{2} \right)^2-t^2=\frac{85}{4}$

$\left(n-\frac{9}{2}+t \right)\left(n-\frac{9}{2} -t\right)=\frac{85}{4}$

$\left(2n-9+2t \right)\left(2n-9 -2t\right)=85$

 2n+2t-9 1 5 17 85 -1 -5 2n-2t-9 85 17 5 1 -85 -17

 n 26 10 10 26 -17 -1 t -21 -3 3 21 21 3

## 延伸討論

[問題] 若 n 為正整數，且 $\sqrt{\frac{9}{4}n^2-3n-2}$ 也是正整數，請問 n = ?

[解答]

$\frac{9}{4}n^2-3n-2=t^2$

$\frac{9}{4}\left[n^2-\frac{4}{3}n+{\left( \frac{2}{3}\right)}^{2} \right]-2=t^2+\frac{9}{4}\left( {\frac{2}{3}}\right)^{2}$

$\frac{9}{4}\left[n-\frac{2}{3} \right]^2-2=t^2+1$

$\frac{9}{4}\left[n-\frac{2}{3} \right]^2-t^2=3$

$\left[\frac{3}{2}n-1 \right]^2-t^2=3$

$\left[\frac{3}{2}n-1 +t\right]\left[\frac{3}{2}n-1 -t\right]=3$

$(3n-2+2t)(3n-2-2t)=12$

 3n-2+2t 12 6 4 3n-2-2t 1 2 3

 3n+2t 14 8 6 3n-2t 3 4 5

 n 17/6 2 11/6 t 11/4 1 1/4