A224241 Numbers k such that k^2 XOR (k+1)^2 is a square, and k^2 XOR (k+2)^2 is also a square, where XOR is the bitwise logical exclusive-or operator.

0, 3, 130456, 342096, 1226720, 291575011, 379894587, 523040160, 15216609776, 136622606520

Offset: 1

Author

Alex Ratushnyak, Apr 01 2013

Keywords

• nonn
• base
• more
• less

Comments

A subsequence of A221643.

Conjecture: the sequence is infinite.

Crossrefs

Cf. A221643.

Programs

C

#include 
#include 
int main() {
  unsigned long long a, i, t;
  for (i=0; i < (1L<<32)-2; ++i) {
      a = (i*i) ^ ((i+1)*(i+1));
      t = sqrt(a);
      if (a != t*t) continue;
      a = (i*i) ^ ((i+2)*(i+2));
      t = sqrt(a);
      if (a != t*t) continue;
      printf("%llu, ", i);
  }
  return 0;
}

Java

class A224241 {
        static public BigInteger isqrt(final BigInteger n)
        {
                if ( n.compareTo(BigInteger.ZERO) < 0 )
                        throw new ArithmeticException("Negative argument "+ n.toString()) ;
                BigInteger x  ;
                final int bl = n.bitLength() ;
                if ( bl > 120)
                        x = n.shiftRight(bl/2-1) ;
                else
                {
                        final double resul= Math.sqrt(n.doubleValue()) ;
                        x = new BigInteger(""+Math.round(resul)) ;
                }
                final BigInteger two = new BigInteger("2") ;
                while ( true)
                {
                        BigInteger x2 = x.pow(2) ;
                        BigInteger xplus2 = x.add(BigInteger.ONE).pow(2) ;
                        if ( x2.compareTo(n) <= 0 && xplus2.compareTo(n) > 0)
                                return x ;
                        xplus2 = xplus2.subtract(x.shiftLeft(2)) ;
                        if ( xplus2.compareTo(n) <= 0 && x2.compareTo(n) > 0)
                                return x.subtract(BigInteger.ONE) ;
                        xplus2 = x2.subtract(n).divide(x).divide(two) ;
                        x = x.subtract(xplus2) ;
                }
        }
    static public void main(String[] argv)
    {
        for(BigInteger k = BigInteger.ZERO ;  ; k= k.add(BigInteger.ONE) )
        {
            final BigInteger k2 = k.pow(2) ;
            final BigInteger kplus1 = k.add(BigInteger.ONE) ;
            final BigInteger k12 = kplus1.pow(2) ;
            final BigInteger xor1 = k2.xor(k12) ;
            final BigInteger roo1 = isqrt(xor1) ;
            if ( roo1.pow(2).compareTo(xor1) == 0 )
            {
                final BigInteger k22 = kplus1.add(BigInteger.ONE).pow(2) ;
                final BigInteger xor2 = k2.xor(k22) ;
                final BigInteger roo2 = isqrt(xor2) ;
                if ( roo2.pow(2).compareTo(xor2) == 0 )
                    System.out.println(k) ;
            }
        }
    }
}
// R. J. Mathar, Apr 25 2013