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Primitive solutions c to Diophantine equation a^2 + b^3 = c^5, with a, b, c > 0.
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%I #12 Jun 28 2013 13:31:13

%S 8,19,24,28,32,36,75,81,88,96,136,176,224,225,250,328,369,395,432,468,

%T 500,512,537,648,701,710,864,980,1000,1078,1089,1125,1216,1225,1296,

%U 1440,1536,1620,1734,1764,1792,1800,1944,2000,2028,2048,2178,2304,2528,2628

%N Primitive solutions c to Diophantine equation a^2 + b^3 = c^5, with a, b, c > 0.

%C Terms in A178130 not equal to (some previous term)*k^6, k>1.

%C Some c have more than one solutions for (a,b): 432, 1944, 3249, 3528, 5184, 7220, 10000. For example, 432^5 = 3732480^2 + 10368^3 = 3359232^2 + 15552^3, 3528^5 = 714208320^2 + 331632^3 = 464679936^2 + 691488^3 (are there 3 or more solutions?).

%C First two primes are a(2) = 19, a(25) = 701.

%H Zak Seidov, <a href="/A227029/b227029.txt">Table of n, a(n) for n = 1..111</a> (all terms up to 20000)

%e 8^5 = 104^2 + 28^3.

%Y Cf. A178130.

%K nonn

%O 1,1

%A _Zak Seidov_, Jun 28 2013