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Home/ Questions/Q 6739661
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T11:30:55+00:00 2026-05-26T11:30:55+00:00

I have a function Java which looks like this: private static long fieldPrime =

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I have a function Java which looks like this:

private static long fieldPrime = 4294967291L; // 2^32-5
public final static long modMult(long a, long b) {
    long r = (a*b) % fieldPrime;        
    return r;
}

It multiplies two values (which are guaranteed to be between 0 and 2^32-5) and then does modulo a large prime.

It works for most numbers but sometimes a*b overflows and this causes the function to return the wrong value. Unfortunately Java doesn’t support unsigned longs (which would have solved the issue) and BigInteger is too slow. Can I solve this some other way? I.e. can I adjust r somehow when I detect overflow (in this case a*b < 0 always means it has overflowed).

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  1. Editorial Team
    Editorial Team
    2026-05-26T11:30:55+00:00Added an answer on May 26, 2026 at 11:30 am

    This should work (if both a and b are between 0 and fieldPrime-1):

      private static long fieldPrime = 4294967291L; // 2^32-5
      private static long correctionFactor = fieldPrime+25; //fieldPrime + (2^64) mod fieldPrime 
    
      public final static long modMult(long a, long b) {
          long r = (a*b);
          if (r>=0)
          {
            return r % fieldPrime;
          }
          else
          {
            return ((r% fieldPrime)+correctionFactor)%fieldPrime;  
          }
      }
    

    When overflow occurs a*b will actually be a * b – 2^64 so adding (2^64 mod fieldPrime) is what is needed. Adding one more fieldPrime and one more % operation is needed to make the result in range 0 to fieldPrime-1 (otherwise it may be negative).

    (It won’t work this way if fieldPrime>2^32.)

    EDIT
    The else part can also be changed to:

        return (fieldPrime-a)*(fieldPrime-b)%fieldPrime;
    

    (I don’t known which is faster.)

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