A300565 Numbers z such that there is a solution to x^3 + y^4 = z^5 with x, y, z >= 1.

32, 250, 1944, 2744, 3888, 19208, 27648, 55296, 59049, 59582, 81000, 82944, 131072, 135000, 185193, 200000

Offset: 1

Author

M. F. Hasler, Apr 16 2018

Keywords

• nonn
• hard
• more

Comments

Consider a solution (x, y, z), x^3 + y^4 = z^5. For any m, (x*m^20, y*m^15, z*m^12) will also be a solution. If (x/m^20, y/m^15, z/m^12) is a triple of integers, it is also a solution. A solution is called primitive if there is no such m > 1.

If S = a^3 + b^8/4 is a square, for some a,b > 0, then z = b^4/2 + sqrt(S) is in the sequence, with x = a*z and y = b*z. All known terms are of this form, with b in {2, 6, 7, 9, 12}, only for a(2) and a(10) one must consider half-integral b = 5/2 resp. 31/2. Also of this form is z = 81000, 82944, 131072, 135000, 185193, 200000, 243000, 395307, 474552, 574992, 800000, 820125, 862488, 864000, 972000, ... (with integer b), and 444528 (with b = 33/2).

Examples

			a(1) = 32 = 2^5 is in the sequence because (2^5)^5 = (2^6)^4 + (2^8)^3, using 1 + 1 = 2.
a(2) = 250 = 2*5^3 is in the sequence because 250^5 = 2^5*5^15 = (5^4)^4 + (3*5^5)^3, using 5 + 3^3 = 2^5. This solution is special because x and y are not multiples of z.
a(3) = 1944 = 2^3*3^5 is in the sequence because 1944^5 = (2^4*3^6)^4 + (2^5*3^8)^3, using 2 + 1 = 3.
a(7) = 27648 = 2^10*3^3 is in the sequence because (2^10*3^3)^5 = (2^12*3^4)^4 + (2^16*3^5)^3, using 3 + 1 = 2^2.
a(10) = 59582 = 2*31^3 is in the sequence because (2*31^3)^5 = (31^4)^4 + (31^5)^3, using 31 + 1 = 2^5. This is the second case where x and y are not multiples of z.
		

Crossrefs

Cf. A300564 (z^4 = x^2 + y^3), A111925 (z^5 = x^2 + y^4), A302174, A300566 (z^6 = x^4 + y^5).

Programs

PARI

is(z)=for(y=1,sqrtnint(-1+z=z^5,4),ispower(z-y^4,3)&&return(y))

Extensions

a(11)-a(16) from Giovanni Resta, Jan 23 2020