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A300566 Numbers z such that there is a solution to x^4 + y^5 = z^6 with x, y, z >= 1. 9

%I #41 Jan 19 2019 04:15:43

%S 8748,10368,342732

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

%C 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

%C 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

%C Observation: a(n) = A054744(n+38) = A257999(n+32), at least for 1 <= n <= 2 in both cases. - _Omar E. Pol_, Apr 17 2018

%C 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 non-primitive terms in A054744. - _M. F. Hasler_, Apr 19 2018

%C 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

%e 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^(10k-3) are in the sequence for k >= 1, with x = 8*3^(15k-5) and y = 8*3^(12k-4).

%e 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).

%e 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.

%e 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).

%e 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.

%o (PARI) is(z)=for(y=1,sqrtnint(-1+z=z^6,5),ispower(z-y^5,4)&&return(y))

%o /* Code below for illustration only, not guaranteed to give a complete list. Half-integral values give the additional term 31505922 for b = 63/2. Third-integral values give the additional solution z = 342732 for b = 26/3. */

%o 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

%Y Cf. A300564 (z^4 = x^2 + y^3) and A242183, A300565 (z^5 = x^3 + y^4), A302174.

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

%Y See A303266 for the y-values.

%K nonn,more,bref,hard

%O 1,1

%A _M. F. Hasler_, Apr 16 2018

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