Question

n(n+1)(n+2)(n+3)+1

n是正整数，求证一定是完全平方数

Answer 1

原式=n(n+3)(n+1)(n+2)+1 =(n^2+3n)(n^2+3n+2)+1 设n^2+3n=a 所以原式=a(a+2)+1 =a^2+2a+1 =(a+1)^2 =(n^2+3n+1)^2 所以原式一定是一个完全平方式 o(∩_∩)o...

Answer 2

把n(n+1)(n+2)(n+3)+1算出来,结果是n^4+6n^3+11n^2+6n+1
设
(n2+bn+c)的平方等于上面那个数
则
c^2=1
2b=6
2c+b^2=11
2bc=6
解得
b=3
c=1
所以(n^2+3n+1)^2=n(n+1)(n+2)(n+3)+1

Answer 3

把(n+1)(n+2)(n+3)(n+4)+1算出来,结果是n^4+10n^3+35n^2+50n+25
设
(n2+bn+c)的平方等于上面那个数
则
c^2=25
2b=10
2c+b^2=35
解得
b=5
c=5
所以是n^2+5n+5