Method in Java that returns the first odd found in a range -

i need create method receives int n , tests integers between n*n+1 , (n+1)*(n+1)-1 , returns first odd number encounters. if no such number found between these bounds function return 0.

i'm trying figure out how while , don't it. i'm quite new java.

can me?

i thought in eclipse says me this method must return result of type int , didn't understand why.

public static int test(int n){      // receives argument n          int = n*n+1;             // calculate lower bound         int b = (n+1)*(n+1)-1;     // calculate upper bound           {          (int = 1; <=a; i++){             if (a % == 0){                 return a;             } else if (a % != 0){                 return 0;             }         }          ++a;      } while (a <= b);      }  

to check whether number odd, use modulo:

n % 2 == 0 

if above condition evaluates true, number even. evaluate true if number odd:

n % 2 != 0 

i hope helps.