The tickets

The tickets to the show cost some integer number greater than 1. Also, the sum of the price of the children's and adult tickets, as well as their product, was the power of the prime number. Find all possible ticket prices.

Result

a1 =  2
a2 =  4
a3 =  8
a4 =  16

Solution:

 a+b=p1x a b=p2y  p1,p2...prime n2=2,4,6,8....,2k  a=b=p2  a+b=2 p2=p1x =>p1=2 p1 p2=p1x=0  a1=2 b1=2  s1=2+2=4=22 s2=2 2=4=22 \ \\ a+b = p_{ 1 }^x \ \\ a \cdot \ b = p_{ 2 }^y \ \\ \ \\ p_{ 1 }, p_{ 2 } ... prime \ \\ n_{ 2 } = 2,4,6,8...., 2k \ \\ \ \\ a = b = p_{ 2 } \ \\ \ \\ a+b = 2 \ p_{ 2 } = p_{ 1 }^x \ \\ = > p_{ 1 } = 2 \ \\ p_{ 1 } \cdot \ p_{ 2 } = p_{ 1 }^x = 0 \ \\ \ \\ a_{ 1 } = 2 \ \\ b_{ 1 } = 2 \ \\ \ \\ s_{ 1 } = 2+2 = 4 = 2^2 \ \\ s_{ 2 } = 2 \cdot \ 2 = 4 = 2^2
a2=4 b2=4  s1=4+4=8=23 s2=4 4=16=24a_{ 2 } = 4 \ \\ b_{ 2 } = 4 \ \\ \ \\ s_{ 1 } = 4+4 = 8 = 2^3 \ \\ s_{ 2 } = 4 \cdot \ 4 = 16 = 2^4
a3=8 b3=8  s1=8+8=16=24 s2=8 8=64=26a_{ 3 } = 8 \ \\ b_{ 3 } = 8 \ \\ \ \\ s_{ 1 } = 8+8 = 16 = 2^4 \ \\ s_{ 2 } = 8 \cdot \ 8 = 64 = 2^6
a4=16 b4=16  s1=16+16=32=25 s2=16 16=256=28a_{ 4 } = 16 \ \\ b_{ 4 } = 16 \ \\ \ \\ s_{ 1 } = 16+16 = 32 = 2^5 \ \\ s_{ 2 } = 16 \cdot \ 16 = 256 = 2^8







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