24/04/12 21:22:50.96 W3OozUMf.net
>>133,134
(1)
3^n = k^3 + 1
= (k+1)(kk－k+1)
= (k+1){(k+1)^2－3(k+1) + 3},
∴ k+1 = 3^{p+1}, (p≧0)
(右辺) = 3^{p+1} (3^{p+2}(3^p－1) + 3) … (A)
(A) が3の累乗で表わせるためには
3^p－1 = 0,
p = 0,
k = 2,
n = 2.
(2)
(-1)^n ≡ 3^n = kk－40 ≠ -1 (mod 4)
∴ n = 2m, (偶数)
∴ －40 = 3^n－kk = (3^m +k)(3^m -k),
3^m ≦ 40－k < 40 より
m = 1, 2, 3,
n = 2. 4. 6,
k = 7, 11, なし.