Factorization: - 45x ^ 4Y ^ 2Z ^ 3 + 18x ^ 3Y ^ 2Z ^ 2 + 90x ^ 2Y ^ 2Z ^ 2

Factorization: - 45x ^ 4Y ^ 2Z ^ 3 + 18x ^ 3Y ^ 2Z ^ 2 + 90x ^ 2Y ^ 2Z ^ 2

-45x^4y^2z^3+18x^3y^2z^2+90x^2y^2z^2
=-9x²y²z²(5x²z-2x-10）

【1】 2X + 3Y + 4Z = 5 [2] x-2y + 3Z = 8 find 8x + 5Y + 18z

2x+3y+4z=5 (1)
x-2y+3z=8 (2)
(1)×3+(2)×2
6x+9y+12z+2x-4y+6z=15+16
8x+5y+18z=31

X + y + Z + XY + XZ + YZ + XYZ = 182 (x, y, Z are all positive integers, and x > y > z) to find x, y, Z

∵x+y+z+xy+xz+yz+xyz=182
∴1+x+y+z+xy+xz+yz+xyz=183
∴（1+x)(1+y)(1+z)=1*3*61
Isn't that right
x+1=1,y+1=3,z+1=61
x=0,y=2,z=60

X + y + Z + XY + XZ + YZ + XYZ = 181 (x, y, Z are all positive integers, and x > y > z) to find x, y, Z

The original form is arranged as
(x+1)(y+1)(z+1)=182
182=2*7*13
X, Z, y are 1 612

If x + y + Z = XYZ (x, Z, Y > 0), find the maximum value of 1 / (1 + XY) + 1 / (1 + XZ) + 1 / (1 + YZ)

Obviously XY > 1. XZ > 1. YZ > 1
simple form

A and B are prime numbers. A + B is less than 100 and a multiple of 7. If a + B is odd, what is AXB
Hurry, hurry, hurry, hurry, hurry

19 and 2

I've worked out the formula for prime number. Let's check it: the power of n of 6 minus the power of P equals prime number (n is sum number, P is prime number)
I've got 30, and I'm not wrong,

Sorry, your conclusion is not correct. Let n = 75, P = 11, then your expression can be divided by 19. The proof is as follows:
6 ^ 5 ≡ 5 (mod19), by Fermat's theorem, 6 ^ 18 ≡ 1 (mod19), thus [(6 ^ 75) ^ 11] ≡ 6 ^ 5 * [(6 ^ 18) ^ 15] ≡ 5 (mod19), so [(6 ^ 75) ^ 11] - 5 can be divided by 19, not prime
By the way, the prime number formula is a problem that mathematicians have been interested in since ancient times. Fermat once made a conjecture, which was later overthrown by Bangla. So far, no formula can always produce prime numbers

Let p be a prime number greater than 5, and prove that the fourth power of P ≡ 1 (mod24)

P ^ 4-1 = (P ^ 2 + 1) (P + 1) (p-1), because P is a prime number greater than 5, so p + 1, P-1 are two continuous even numbers, so one of them must be a multiple of 4, and the other is a multiple of 2. So 8 | (P + 1) (p-1)
On the other hand, P + 1, P, P-1 are three continuous positive integers, so the remainder of module 3 is different, so if P-1 and P + 1 are not multiples of three, then p is a multiple of 3, but p is a prime number greater than 5, which does not contain the factor of 3, so 3 | (P + 1) (p-1), and (3,8) = 1, so 24 | (P + 1) (p-1), so 24 | (P + 1) (P ^ 2 + 1), that is 24 | p ^ 4-1
So p ^ 4 ≡ 1 (mod24) is proved

Why is it not a prime number to add 1 to the (5th power) power of 2

Just ask: is there any way to prove it?

Prove that the 1984 power + 1 of 2 is not prime

1984 = 64 * 31, it's easy to have an odd number here, because x ^ n + 1 can be decomposed when n is an odd number
2^1984+1
=(2^64)^31+1
=(x^64+1)(.）
The bracket on the right is omitted, because it has been explained that 1984 power + 1 can be decomposed into the product of two factors, so it is a composite number