

A300566


Numbers z such that there is a solution to x^4 + y^5 = z^6 with x, y, z >= 1.


9




OFFSET

1,1


COMMENTS

Also in the sequence: 810000 = 2^4*3^4*5^4, 1361367 = 3^4*7^5, 3240000 = 2^6*3^4*5^4, 9335088 = 2^4*3^5*7^4, 25312500 = 2^2*3^4*5^7, 31505922 = 2*3^8*7^4, 43740000 = 2^5*3^7*5^4, 512578125 = 3^8*5^7, 1215000000 = 2^6*3^5*5^7, 1701708750 = 2*3^4*5^4*7^5, 2196150000 = 2^4*3*5^5*11^4, 2431012500 = 2^2*3^4*5^5*7^4, 4269246912 = 2^6*3^4*7^7, 4447203750 = 2*3^5*5^4*11^4, 36015000000 = 2^6*3*5^7*7^4, 48717927500 = 2^2*5^4*11^7, 75969140625 = 3^4*5^8*7^4, 91116682272 = 2^5*3^4*7^4*11^4.  Jacques Tramu, Apr 17 2018
Consider a solution (x,y,z) of x^4 + y^5 = z^6. For any m, (x*m^15, y*m^12, z*m^10) is also a solution. Reciprocally, if (x/m^15, y/m^12, z/m^10) is a triple of integers for some m, then this is also a solution. We call primitive a solution for which there is no such m > 1.  M. F. Hasler, Apr 17 2018
These relations hold only for n = 1 and 2. The next larger known term 342732 = 2^2*3*13^4 shows that in general the terms don't belong to A054744 nor A257999, although the earlier comment implies that each term gives rise to infinitely many nonprimitive terms in A054744.  M. F. Hasler, Apr 19 2018
When S = a^4 + b^10/4 is a square, then z = b^5/2 + sqrt(S) is a solution, with x = a*z and y = b*z. All known solutions and further solutions 8957952, 10616832, 52200625, 216486432, ... are of this form (with rational a, b).  M. F. Hasler, Apr 19 2018


LINKS



EXAMPLE

a(1) = 8748 = 2^2*3^7 is in the sequence because 8748^6 = (2^3*3^8)^5 + (2^3*3^10)^4, using 2^3 + 1 = 3^2. Similarly, all z = 4*3^(10k3) are in the sequence for k >= 1, with x = 8*3^(15k5) and y = 8*3^(12k4).
a(2) = 10368 = 2^7*3^4 is in the sequence because 10368^6 = (2^8*3^5)^5 + (2^10*3^6)^4, using 3 + 1 = 2^2. Similarly, any z = 2^7*3^(10k+4) is in the sequence for k >= 0, with x = 2^10*3^(15k+6) and y = 2^8*3^(12k+5).
z = 342732 = 2^2*3*13^4 is in the sequence because (2^2*3*13^4)^6 = (2^3*13^5)^5 + (2^3*5*13^6)^4, using 2^3*13 + 5^4 = 3^6.
z = 810000 = 2^4*3^4*5^4 is in the sequence because z^6 = x^4 + y^5 with x = 2^5*3^6*5^6 and y = 2^4*3^5*5^5 (using 1 + 3*5 = 2^4).
z = 1361367 = 3^4*7^5 is in the sequence because z^6 = x^4 + y^5 with x = 3^5*7^8 and y = 2*3^4*7^6.


PROG

(PARI) is(z)=for(y=1, sqrtnint(1+z=z^6, 5), ispower(zy^5, 4)&&return(y))
/* Code below for illustration only, not guaranteed to give a complete list. Halfintegral values give the additional term 31505922 for b = 63/2. Thirdintegral values give the additional solution z = 342732 for b = 26/3. */
S=[]; N=1e5; forstep(b=1, 9, 1/3, forstep(a=1, N, 1/3, issquare(b^10+a^4<<2, &r)&& !frac(z=b^5/2+r/2)&& !print1(z", ")&&S=setunion(S, [z])); print1([b])); S


CROSSREFS

Cf. A100294: numbers of the form a^5 + b^4.


KEYWORD

nonn,more,bref,hard


AUTHOR



STATUS

approved



