Mathematical proof problem: when n > 2, the nth power of any prime number cannot be decomposed into the sum of the same power (nth power) of two prime numbers Because a / (B + 2) = 1, B = A-2, why

Suppose there are three prime numbers a, B, C, a > b > C
Suppose a ^ n = B ^ n + C ^ n (n is larger than 2 and odd)
A ^ n = (B + C) * f (B, c) (f (B, c) is a polynomial)
f(b,c)= a^n/(b+c)
It must be an integer B, C, f (c)
If B and C are prime numbers greater than 2, it is obvious that B + C is even,
However, a ^ n / (B + C) is an integer and a is odd (a obviously cannot be equal to the smallest prime 2)
If one of B and C is 2
Then a = a, B = B, C = 2
A ^ n = (B + 2) * f (B, 2) because a / (B + 2) = 1, B = A-2
A ^ n - (A-2) ^ n-2 ^ n is obviously not zero, so the hypothesis is not true
Still assume a ^ n = B ^ n + C ^ n (n is larger than 2 and even)
C ^ n = (a-b) * g (a, b) g is a polynomial
Similarly, A-B is even. When C is not 2, the hypothesis does not hold
When C = 2, 2 ^ n = (a-b) * g (a, b)
Because 2 / (a-b) = 1, B = A-2
There is still a ^ n-2 ^ n - (A-2) ^ n which is obviously not zero
It seems a little troublesome, who can give a simple study
Therefore, it has been proved

If (x + 3Y / 4x-y) = 3, then (2x / y) =?

12/11

According to (x + 12) - 27 (x is an integer), when x = (), the number obtained from the above formula is the smallest prime number

The smallest prime number is 2
So (x + 12) - 27 = 2
x+12=29
x=17
If it helps you, please remember to adopt it_ Thank you

When 27 is divided into the sum of several different prime numbers, to maximize the product of the prime numbers, these prime numbers are______ ．

According to the analysis, these prime numbers are 2, 5, 7 and 13 respectively, so the answer is: 2, 5, 7 and 13

If you want to make the product of these prime numbers maximum, what are the prime numbers?

Prime numbers less than 33 are arranged from small to large: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 (a total of 11). Because 2 + 3 + 5 + 7 + 11 < 33, and 2 + 3 + 5 + 7 + 11 + 13 > 37, they can be divided into the sum of five different prime numbers at most. However, because 33 is odd, there can be no even prime number 2 among the five different prime numbers removed, otherwise the sum of the other four odd prime numbers is even, and the sum of these five prime numbers is even and cannot be removed It can be equal to odd number 33, and 3 + 5 + 7 + 11 + 13 = 39 ＞ 33. Therefore, it can be divided into the sum of four different prime numbers at most. To maximize the product of these prime numbers, we must split the prime numbers as large as possible, because 2 + 3 + 5 + 7 + 11 = 28, which is worse than 33: 33-28 = 5; and because 3 + 5 = 8 in 2 + 3 + 5 + 7 + 11, which is just the difference of 5, forms 8 + 5 = 13, so when 33 is decomposed into 2, 7, 11 and 13, the product of prime numbers is the largest These prime numbers are 2, 7, 11 and 13

If you want to make the product of these prime numbers maximum, what are the prime numbers

Divided into 2, 3, 13, 17, the largest product is 1326

36 into the product of two prime numbers
It's OK to multiply a few prime numbers by 36

I can't do it
36=2×2×3×3

2x-{-3y+[4x-(2y-x)]}

There is no way to solve the problem, just go to the bracket layer by layer!
2x-{-3y+[4x-(2y-x)]}
=2x+3y-[4x-(2y-x)]
=2x+3y-4x+(2y-x)
=2x+3y-4x+2y-x
=5y-3x

The solution of the equation a (x + m) 2 + B = 0 of X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), then the solution of the equation a (x + m + 2) 2 + B = 0 is______ ．

∵ the solution of the equation a (x + m) 2 + B = 0 about X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), ∵ the equation a (x + m + 2) 2 + B = 0 is transformed into a [(x + 2) + M] 2 + B = 0, that is, in this equation x + 2 = - 2 or x + 2 = 1, the solution is x = - 4 or x = - 1. So the answer is: X3 = - 4, X4 = - 1

The solution of the equation a (x + m) ^ 2 + B = 0 of X is X1 = - 2, X2 = 1 (a, m, B are all constants to find the value of M)
Don't copy, because it's different from the problem on the Internet. It's to find the value of M. remember to write the process

Solving equation a (x + m) ^ 2 + B = 0
Is ax & # 178; + 2amx + am & # 178; + B = 0
The solution of the equation AX & # 178; + 2amx + am & # 178; + B = 0 is X1 = - 2, X2 = 1,
The quadratic equation a (x + 2) (x-1) = 0 is constructed by X1 = - 2, X2 = 1
That is to say, ax & # 178; + ax-2a = 0
By comparing the coefficients of ax & # 178; + ax-2a = 0 and ax & # 178; + 2amx + am & # 178; + B = 0, it is concluded that
2am=a
M = 1 / 2

Mathematical proof problem: when n > 2, the nth power of any prime number cannot be decomposed into the sum of the same power (nth power) of two prime numbers Because a / (B + 2) = 1, B = A-2, why