A050801 Numbers k such that k^2 is expressible as the sum of two positive cubes in at least one way.

3, 4, 24, 32, 81, 98, 108, 168, 192, 228, 256, 312, 375, 500, 525, 588, 648, 671, 784, 847, 864, 1014, 1029, 1183, 1225, 1261, 1323, 1344, 1372, 1536, 1824, 2048, 2187, 2496, 2646, 2888, 2916, 3000, 3993, 4000, 4200, 4225, 4536, 4563, 4644, 4704, 5184, 5324

Offset: 1

Author

Patrick De Geest, Sep 15 1999

Keywords

• nonn
• nice

Comments

Analogous solutions exist for the sum of two identical cubes z^2 = 2*r^3 (e.g., 864^2 = 2*72^3). Values of 'z' are the terms in A033430, values of 'r' are the terms in A001105.

First term whose square can be expressed in two ways is 77976; 77976^2 = 228^3 + 1824^3 = 1026^3 + 1710^3. - Jud McCranie

First term whose square can be expressed in three ways is 3343221000; 3343221000^2 = 279300^3 + 2234400^3 = 790020^3 + 2202480^3 = 1256850^3 + 2094750^3.

First term whose square can be expressed in four ways <= 42794271007595289; 42794271007595289^2 = 14385864402^3 + 122279847417^3 = 55172161278^3 + 118485773289^3 = 64117642953^3 + 116169722214^3 = 96704977369^3 + 97504192058^3.

First term whose square can be expressed in five ways <= 47155572445935012696000; 47155572445935012696000^2 = 94405759361550^3 + 1305070263601650^3 = 374224408544280^3 + 1294899176535720^3 = 727959282778000^3 + 1224915311765600^3 = 857010857812200^3 + 1168192425418200^3 = 1009237516560000^3 + 1061381454915600^3.

After a(1) = 3 this is always composite, because factorization of the polynomial a^3 + b^3 into irreducible components over Z is a^3 + b^3 = (b+a)*(b^2 - ab + b^2). They may be semiprimes, as with 671 = 11 * 61, and 1261 = 13 * 97. The numbers can be powers in various ways, as with 32 = 2^5, 81 = 3^4, 256 = 2^8, 784 = 2^4 * 7^2 , 1225 = 5^2 * 7^2, and 2187 = 3^7. - Jonathan Vos Post, Feb 05 2011

If n is a term then n*b^3 is also a term for any b, e.g., 3 is a term hence 3*2^3 = 24, 3*3^3 = 81 and also 3*4^3 = 192 are terms. Sequence of primitive terms may be of interest. - Zak Seidov, Dec 11 2013

First noncubefree primitive term is 168 = 21*2^3 (21 is not a term of the sequence). - Zak Seidov, Dec 16 2013

From XU Pingya, Apr 10 2021: (Start)

Every triple (a, b, c) (with a^2 = b^3 + c^3) can produce a nontrivial parametric solution (x, y, z) of the Diophantine equation x^2 + y^3 + z^3 = d^4.

For example, to (1183, 65, 104), there is such a solution (d^2 - (26968032*d)*t^3 + 1183*8232^3*t^6, (376*d)*t - 65*8232^2*t^4, (92*d)*t - 104*8232^2*t^4).

To (77976, 228, 1824), there is (d^2 - (272916*d)*t^3 + 77976*57^3*t^6, (52*d)*t - 228*57^2*t^4, (74*d)*t - 1824*57^2*t^4).

Or to (77976, 1026, 1719), there is (d^2 - (25992*d)*t^3 + 77976*19^3*t^6, (37*d)*t - 1026*19^2*t^4, (11*d)*t - 1710*19^2*t^4). (End)

Examples

			1183^2 = 65^3 + 104^3.
		

References

• Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..612 from T. D. Noe, terms 613..1000 from Harry J. Smith)

Crossrefs

Cf. A050802, A000404, A033430, A001105, A038597, A050803, A106265, A217248.

Programs

Maple

A050801 := proc(n)
    option remember ;
    local a,x,y ;
    if n =1 then
        3
    else
        for a from procname(n-1)+1 do
            for x from 1 do
                if x^3 >= a^2 then
                    break ;
                end if;
                for y from 1 to x do
                    if x^3+y^3 = a^2 then
                        return a ;
                    end if;
                end do:
            end do:
        end do:
    end if;
end proc:
seq(A050801(n),n=1..20) ; # R. J. Mathar, Jan 22 2025

Mathematica

Select[Range[5350], Reduce[0 < x <= y && #^2 == x^3 + y^3, {x,y}, Integers] =!= False &] (* Jean-François Alcover, Mar 30 2011 *)
Sqrt[#]&/@Union[Select[Total/@(Tuples[Range[500],2]^3),IntegerQ[ Sqrt[ #]]&]] (* Harvey P. Dale, Mar 06 2012 *)
Select[Range@ 5400, Length@ DeleteCases[PowersRepresentations[#^2, 2, 3], w_ /; Times @@ w == 0] > 0 &] (* Michael De Vlieger, May 20 2017 *)

PARI

is(n)=my(N=n^2); for(k=sqrtnint(N\2,3),sqrtnint(N-1,3), if(ispower(N-k^3,3), return(n>1))); 0 \\ Charles R Greathouse IV, Dec 13 2013

Formula

a(n) = sqrt(A050802(n)). - Jonathan Sondow, Oct 28 2013

Extensions

More terms from Michel ten Voorde and Jud McCranie