N is a natural number greater than 1. Is it possible for the factorial of n to be a complete square? How to prove the conclusion? If not, is there a series of continuous natural numbers whose product is a complete square?

N is a natural number greater than 1. Is it possible for the factorial of n to be a complete square? How to prove the conclusion? If not, is there a series of continuous natural numbers whose product is a complete square?

No,
Because continuous natural numbers; impossible to find
A1, a1 + 1, a1 + 2; this is an arithmetic sequence, impossible to achieve
Any number is obtained by multiplication of prime numbers. Once a prime number appears in the middle, it is not allowed to find 2 or 3 times of the prime number. It is still a prime number. It is impossible to make the number of prime numbers 2 times of n (natural number)
Factorial of sum 1-20 in MATLAB
>> factorial(20)
ans =
2.4329e+18
Let f (x) = - 1 / 3x3 + 2ax2-3a2x + B (0 ＜ a ＜ 1). When a = 2 / 3, the equation f (x) = 0 of X always has two different real roots in the interval [1,3]. The value range of real number B is obtained
A = 2 / 3, f (x) = - 1 / 3 x ^ 3 + 4 / 3 x ^ 2-4 / 3 x + BF '(x) = - x ^ 2 + 8x / 3-4 / 3 = - 1 / 3 * (3x ^ 2-8x + 4) = - 1 / 3 * (3x-2) (X-2) get the extremum x = 2 / 3, 2 minimum f (2 / 3) = - 8 / 81 + 1 / 27-8 / 9 + B = - 77 / 81 + B maximum f (2) = - 8 / 3 + 16 / 3-8 / 3 + B = B and f (1) = - 1 / 3 + 4 / 3-4 / 3 + B = - 1 / 3 + BF
Let y = f (x) (x ∈ R) satisfy f (x1) + F (x2) = f (x1 * x2) for any real number x1x2, and prove that f (1) = f (- 1) = 0 and f (x) are even functions
f(1)+f(1)=f(1*1)=f(1)
So: F (1) = 0
f(-1)+f(-1)=f((-1)*(-1))=f(1)=0
f(-1)=0
So: F (1) = f (- 1) = 0
f(-x)=f(-1)+f(x)=0+f(x)=f(x)
F (x) is an even function
Let X1 = x2 = 1 and - 1 bring in (x1) + F (x2) = f (x1 * x2) to get f (1) = f (- 1) = 0, let X1 = x, X2 = - 1 bring in (x1) + F (x2) = f (x1 * x2) to get f (x) + F (- 1) = f (- x), that is, f (x) = f (- x), so it is even function
The function f (x) = 1 / 3x3 + 1 / 2ax2 + BX has an extreme point in the interval [- 1,1), (1,3] to find the maximum value of a2-4b
Because f (x) = 1 / 3x3 + 1 / 2ax2 + BX has an extreme point in the interval [- 1,1), (1,3], because the derivative at the extreme point is 0, so the two roots of F '(x) = x ^ 2 + ax + B are in the interval [- 1,1), (1,3], respectively
It is known that the function f (x) = 4 ^ x + 4 ^ (- x) is even. It is proved that for any real number X1 and X2, 1 / 2 [f (x1) + F (x2)] ≥ f [(x1 + x2) / 2]
Prove: 1 / 2 [f (x1) + F (x2)] = 1 / 2 [4 ^ X1 + 4 ^ (- x1) + 4 ^ x2 + 4 ^ (- x2)]
=1 / 2 (4 ^ X1 + 4 ^ x2) + 1 / 2 [4 ^ (- x1) + 4 ^ (- x2)] by means inequality
≥4^[(x1+x2)/2]+4^[-(x1+x2)/2] = f[(x1+x2)/2]
Find the monotone decreasing interval of the function y = 13x3-12 (a + A2) x2 + a3x + A2
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It is known that a and B are real numbers, f (x) = x2 + ax + 1, and f (x + 1) is an even function in the domain of definition. The function g (x) = - BF [f (x + 1)] + (3b-1) f (x + 1) + 2 is an increasing function in (- ∞, - 2) and a decreasing function in (- 2,0),
Finding the value of a and B
F (x + 1) = (x + 1) ^ 2 + a (x + 1) + 1 = x ^ 2 + (2 + a) x + 2 + A is an even function in the domain of definition
2+a=0,a=-2
f(x)=x^2-2x+1=(x-1)^2
f(x+1)=x^2
g(x)=-bf(x^2)+(3b-1)x^2+2=-b(x^2-1)^2+(3b-1)x^2+2
=-bx^4+(5b-1)x^2+2-b
g'(x)=-4bx^3+2(5b-1)x
It is a decreasing function on (- ∞, - 2) and an increasing function on (- 2,0)
G (x) reaches a minimum at x = - 2
g'(-2)=32b-4(5b-1)=0,b=-1/3
a=-2,b=-1/3