1楼一叶轻舟
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发表于 2009-5-6 08:07
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解: n^4 - 4n^3 + 22n^2 - 36n + 18
= n^4 - 4n^3 + 4n^2 + 18n^2 -36n + 18
= (n^2-2n)^2 + 18(n^2-2n) +18
= (n^2-2n+9)^2 - 63
= k^2 (k∈N)
∴(n^2-2n+9)^2 -k^2 =63
即(n^2-2n+9-k)(n^2-2n+9+k) = 63
∵△ = 4-4(9+k) = -32-4k < 0 ∴n^2-2n+9+k > 0 ∴n^2-2n+9-k > 0
(1) n^2-2n+9-k = 1, n^2-2n+9+k =63, 此时无正整数解
(2) n^2-2n+9-k = 3, n^2-2n+9+k =21, 解得 n =3
(3) n^2-2n+9-k = 7, n^2-2n+9+k = 9, 解得 n =1
所以满足条件的正整数解为: n=1和n=3
[ 本帖最后由 一叶轻舟 于 2009-5-6 08:23 编辑 ].
