A(=abcde) is a 5-digit natural number with no leading zeros and e>a. B(=eca), 3-digit natural number is a multiple of 9 while C(=db), 2-digit natural number is a multiple of 3 but not 9. Show that x^{5}-y^{5}=A has no integer solutions if sum of digits of C is not 6.

Note: A(=abcde) means Ten Thousand's digit of A:a, Thousand's digit of A:b and so on unit's digit of A:e.

Possible values a+c+e can take = 9,18
Possible values b+d can take = 3,12,15
So, Possible values that a+c+e-b-d can take=6,-3,-6,15,6,3.
Possible values of {(a+c+e-b-d) mod 11} = 3,4,5,6,8.
A=10^4 *a+10^3 *b+10^2 *c+10d+e
=> A mod 11 = (a-b+c-d+e) mod 11
=> x^5-y^5 mod 11 = {3,4,5,6,8} ---- (1)
Consider x^10 mod 11
If x is relatively prime to 11, then by Fermat's Theorem
x^10 mod 11 =1 => x^5 mod 11 = +/-1.
If x is a multiple of 11, then x^5 mod 11 = 0.
Possible values of x^5 mod 11 ={0,+1,-1}.
Possible values of x^5-y^5 mod 11={0,+1,-1,-2,+2}={0,1,2,9,10} ----(2)
There are no common elements for sets 1 and 2. So, there are no solutions for x^5-y^5 = A.

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