## S000832

Sequence of sorted prime quadruples (a,b,c,d) such that a^3 + b^3 + c^3 + d^3 = 2016.

349, 277, 167, -409, 463, 419, -13, -557, 1123, 571, 149, -1171, 1637, -113, -1181, -1399, 1877, -263, -443, -1867, 2153, 1061, 271, -2237, 3457, -811, -1423, -3359, 3709, 2099, -1259, -3877, 5009, -863, -1709, -4933, 7507, 1889, 1733, -7577

1

All these quadruples have at least one signed term because the only all-positive solutions are {10, 10, 2, 2} and {12, 6, 4, 2}, which clearly not all primes. The rather fast, but memory intensive, program for computing these solutions is given below.

T. D. Noe, Plot of 121 quadruples

T. D. Noe, Table of 121 quadruples

Seiji Tomita, Prime Puzzle 817: 2016 and prime numbers

(Mma) nn = PrimePi[10000]; year = 2016; Clear[a, b]; lim1 = Prime[nn]^3 + 2^3; lim2 = Prime[nn]^3 - 2^3; ts = Reap[Do[q = a^3 + b^3; If[a >= b && q <= lim1, Sow[q]], {a, Prime[Range[nn]]}, {b, Prime[Range[nn]]}]]; ts = Union[ts[[2, 1]]]; td = Reap[Do[q = a^3 - b^3; If[0 < q <= lim2, Sow[q]], {a, Prime[Range[nn]]}, {b, Prime[Range[nn]]}]]; td = Union[td[[2, 1]]]; s1 = Intersection[ts, ts + year]; t1 = Table[{FindInstance[a^3 + b^3 == s1[[i]] && a > 0 && b > 0, {a, b}, Integers], FindInstance[a^3 + b^3 == s1[[i]] - year && a > 0 && b > 0, {a, b}, Integers]}, {i, Length[s1]}]; s2 = Intersection[ts, td + year]; t2 = Table[{FindInstance[a^3 + b^3 == s2[[i]] && a > 0 && b > 0, {a, b}, Integers], FindInstance[a^3 - b^3 == s2[[i]] - year && a > 0 && b > 0, {a, b}, Integers]}, {i, Length[s2]}]; s3 = Intersection[td, ts + year]; t3 = Table[{FindInstance[a^3 + b^3 == s3[[i]] - year && a > 0 && b > 0, {a, b}, Integers], FindInstance[a^3 - b^3 == s3[[i]] && a > 0 && b > 0, {a, b}, Integers]}, {i, Length[s3]}]; u1 = {}; Do[q = Flatten[t1[[i]]]; q1 = {q[[2, 2]], q[[1, 2]], -q[[3, 2]], -q[[4, 2]]}; AppendTo[u1, q1], {i, Length[t1]}]; u2 = {}; Do[q = Flatten[t2[[i]]]; q1 = Flatten[{Reverse[Sort[{q[[1, 2]], q[[2, 2]], q[[4, 2]]}]], -q[[3, 2]]}]; AppendTo[u2, q1], {i, Length[t2]}]; u3 = {}; Do[q = Flatten[t3[[i]]]; q1 = Reverse[Flatten[Sort[-{q[[1, 2]], q[[2, 2]], q[[4, 2]], -q[[3, 2]]}]]]; AppendTo[u3, q1], {i, Length[t3]}]; Union[u1, u2, u3]

sign,nice

T. D. Noe, Jan 24 2016, Jan 28 2016, Feb 01 2016, Feb 04 2016