Check if there exists a triplet such that a ^ n + b ^ n = c ^ n
int pow_sum = pow ( a , n ) + pow ( b , n ) ; double c = pow ( pow_sum , 1.0 / n ) ; int c_pow = pow ( ( int ) c , n ) ; if ( c_pow == pow_sum ) { cout << " Count ▁ example ▁ found " ; return ; } } cout << " No ▁ counter ▁ example ▁ within ▁ given " " ▁ range ▁ and ▁ data " ; }