Find the number of solutions in integers(x,y) of the equation x^2 y^3=6^12

Question

help..pls.. number theory..

Answer 1

Prime-factor the right side: 6¹² = 2¹²3¹²

 

Now we know that x = ±2^a3^b and y = 2^c3^d for some integers a,b,c,d ≥ 0

 

(±2^a3^b)² (2^c3^d)³ 2¹² 3¹²

 

2^2a+3c 3^2b+3d = 2¹² 3¹²

 

2a + 3c = 12

 

2b + d = 12

 

For a and c, the solutions are

 

a = 0, c = 4

 

a = 3, c = 2

 

a = 6, c = 0

 

For b and d, the solution set is the same and independent.

 

So plugging these in gives all the solutions to x²y³ = 6¹².

 

(±1 , 1296)

 

(±27 , 144)

 

(±729 , 16)

 

(±8 , 324)

 

(±216 , 36)

 

(±5832 , 4)

 

(±64 , 81)

 

(±1728 , 9)

 

(±46656, 1)