2022/051) Show tthat there are infnite solutions to $x^3+y^4= z^{31}$

We know that 2^m * 2 is a power of 2

so if we chose $x^3 = y^4 = 2^t$ for some t then we shall get power of 2

so let $x = 2^{4m}$ and $y = 2^{3m}$

so $x^3 + y^ 4= 2^{12m} + 2^{12m} = 2^{12m+1}$

 Now there is only one 1ssure remanins that is 12m + 1 has to be multiple of 31

that is not a problem as gcd(31,12) is 1 and we shall use exteneded euclidean algoirithm to find the same

now 31 =  2 * 12 + 7 we take the highest multiple of 12 and leave the remainder

or $7 = 31 - 2 * 12\cdots(1)$

12 = 7 * 1 + 5

or $5 = 12 - 7 * 1 = 12 - (31 - 2 * 12)$ using (1)

or  $5 = 12 * 3 - 31\cdots(2)$

now $ 7 = 5 + 2$

or $2 =  7- 5 = (31- 2 * 12) - (12 * 3 -31) = 2 * 31 - 5 * 12$ (uinsg (1) and (2)

or $2 = 2 * 31 - 5 * 12$

now $5 = 2 * 2 + 1$

or $1= 5 - 2 * 2$

or $1 = (12 * 3 - 31) - 2 * ( 2 *31 - 5 *12) = 13 * 12 - 5 * 31$

so if m is 13 or any number 31n + 13 we get 12m+ 1 is a multiple of 31

m = 31n + 13 gives 12m +1 = 12(31n + 13) = 31(12n + 5)

so $x = 2^{4(31n + 13)}$, $y = 2^{3(31n + 13)}$ and $z = 2^{12n+5}$ is parametric solution and hence infinite solutions.