三個方程式求其共同解的數學問題？

Question

                                   {yz=x2zx,y,z為自然數且滿足{3y=3×27z，則(x,y,z)=？                                   {x+2y+z=41

Answer 1

y^z=x^(2z)=(x^2)^z
y=x^2

3^y=3×27^z=3^(3z+1)
y=3z+1

解聯立
x+2y+z=41
y=x^2
y=3z+1

得
x=4
y=16
z=5

Answer 2

1.
y^z=x^2z
→y^z=(x^2)^z
因為z不等於0
y=x^2

2.
3^y=3*27^z
→3^y=3^(3z+1)
→y-3z=1
→x^2-3z=1......(1)

3.
x+2y+z=41......(2)
(2)*3+(1)
→7x^2+3x-124=0
x=4 or -31/7(不合)
y=16
z=5
(x,y,z)=(4,16,5)

Answer 3

由yz=x2z知x2=y由3y=3×27z知y=3z+1代入第三式化簡，7x2+3x-124=0，x=4或-31/7(不合)(x,y,z)=(4,16,5)