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.
No comments:
Post a Comment