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Sunday, July 10, 2022

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.



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