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A300565
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Numbers z such that there is a solution to x^3 + y^4 = z^5 with x, y, z >= 1.
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9
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32, 250, 1944, 2744, 3888, 19208, 27648, 55296, 59049, 59582, 81000, 82944, 131072, 135000, 185193, 200000
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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PROG
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(PARI) is(z)=for(y=1, sqrtnint(-1+z=z^5, 4), ispower(z-y^4, 3)&&return(y))
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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