It is proved that the sum of the product of four continuous integers and one must be a complete square number

It is proved that the sum of the product of four continuous integers and one must be a complete square number

Let the four consecutive integers be n, N + 1, N + 2, N + 3 from small to large
n(n+1)(n+2)(n+3)
=(n^2+3n)(n^2+8n+2)
=(n^2+3n+1)^2-1
n(n+1)(n+2)(n+3)+1
=(n^2+3n+1)^2
The product of four consecutive integers plus 1 is a complete square number

Try to explain: than the product of four consecutive positive integers 1 number must be the square of an integer

It is proved that if the four consecutive integers are n, N + 1, N + 2 and N + 3, then
n(n+1)(n+2)(n+3)+1
=n(n+3)(n+1)(n+2)+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2
So the sum of the product of 4 consecutive integers and 1 is a complete square number

Proof: the product of four consecutive integers plus 1 is the square of an integer

Let the four consecutive integers be n-1, N, N + 1, N + 2, then (n-1) n (n + 1) (n + 2) + 1, = [(n-1) (n + 2)] [n (n + 1)] + 1 = (N2 + n-2) (N2 + n) + 1 = (N2 + n) 2-2 (N2 + n) + 1 = (N2 + n-1) 2. Therefore, the product of four consecutive integers plus 1 is the square of an integer

If the solution set of X * 2-2x ≤ a * 2-2a-1 on R is empty, the value range of real number a

Let y = x ^ 2-2x = x (X-2) get the minimum value of - 1 when x = 1
If the solution set is empty, a ^ 2-2a-1

Let a = {x | x ^ 2 + 2x-3} and B = {x | x ^ 2-2a-1 ≤ 0, a > 0}. If a ∩ B contains an integer, then the value range of real number a is

From x ^ 2 ＋ 2x-3 ＞ 0, we can get: (x-1) (x + 3) ＞ 0, X ＜ - 3, or X ＞ 1. A = {x ＜ - 3, or X ＞ 1}. From x ^ 2-2ax-1 ≤ 0, we can get: x ^ 2-2ax + A ^ 2 ≤ 1 + A ^ 2, x-a ^ 2 ≤ 1 + A ^ 2, x-a ^ 2 ≤ 1 + A ^ 2, x-a ^ 2 ≤ x-a ＜ √ (1 + A ^ 2), a - √ (1 + A ^ 2)

Let m = 2A (a)_ 2)+7,n=(a_ 2) (A-3), compare the size of M, n

M-N = 2A ^ 2-4a + 7-a ^ 2 + 5a-6 = a ^ 2 + A + 1 = (a + 1 / 2) ^ 2 + 3 / 4, greater than 0, so m > n

Let m = 2A (A-2), n = (A-1) (A-3), then compare the size of M and n

m-n=2a^2-4a-a^2-3+4a=a^2-3
A = radical 3 M = n
a> Root 3 m > n
A = root 3 M

Let m = (2A (A-2) + 4, n = (A-1) (A-3), then the size relation of M and n

M-N=2a(a-2)+4-(a-1)(a-3)
=2a^2-4a+4-a^2+4a-3
=a^2+1
＞0
So m > n

If M = (2a-1) * (a + 2), n = (a + 2) * (A-3), then the size relation of M and N?

m-n
=(2a-1)*(a+2)-(a+2)(a-3)
=(a+2)[(2a-1)-(a-3)]
=(a+2)(a+2)
=(a+2)^2
m≥n

(a-b) & #178; = (a + b) & #178; - m, then M =?

(a－b）²＝（a＋b)²－M
a²-2ab+b²=a²+2ab+b²-M
M=a²+2ab+b²-（a²-2ab+b²）
M=4ab