=n²(n²-4n+4)+18n²-36n+18
=n²(n-2)²+18n²-36n+18
=(n²-2n)²+18(n²-2n)+18
=(n²-2n)²+18(n²-2n)+81-63
=[(n²-2n)+9]²-63
=(n²-2n+9)²-63
设上式等于k²,k为正整数,则有
(n²-2n+9)²-63=k²
(n²-2n+9)²-k²=63
(n²-2n+9-k)(n²-2n+9+k)=63=1×3×3×7
显然,(n²-2n+9-k)两个题都有了吗？第二题：n⁴+6n³+11n²+3n+31>n⁴+6n³+9n²=(n^2+3n)²n⁴+6n³+11n²+3n+31-(n²+3n+4)²=15-6n²-21n<0所以(n²+3n)²