Please see also: UPDATE 1 UPDATE 2 .

In other words, does the Diophantine equation a^{3} + b^{3} = c^{2}
have any solutions with a,b both prime?

Yes, here are two examples: 11^{3} + 37^{3} = 228^{2}
and 8663^{3} + 2137^{3} = 812340^{2}.

In fact there are lots of solutions with a,b both prime. Ed Pegg, on the sequence fans
mailing list asked the question:

Is the number, c, which occurs in a solution with a,b prime always
divisible by 12?

I gave a fairly complete analysis of this question and emailed it to the list on Saturday, October 23, 2004 5:56 PM. Subsequently, Dean Hickerson send a proof to the sequence fans list, different from mine in that it did not use parameterizations. For some reason (email problems?) my message never appeared on the list. So I sent it again the next day. Here is that message:

Ed Pegg, based on numerical information from Kirk Bresniker, asks whether a,b prime and a^3 + b^3 = c^2 implies that c is divisible by 12. Hugo Pfoertner, Ralf Stephan, an myself provided additional numerical evidence and comments. Finally, Paul C. Leopardi proved a lemma that is almost the desired result, but unless I misunderstood needed an additional hypotheses. Here is additional information which amounts to a proof and some parameterized solutions, but some details omitted. Lemma 1: if gcd(a,b)=1 then gcd(a+b,a^2-a*b+b^2)=1 or 3. proof: Since (a+b)^2-(a^2-a*b+b^2) = 3*a*b, any prime p<>3 which divides both a+b and a^2-a*b+b^2 will divide a or b and a+b hence it will divide both a and b contrary to gcd(a,b)=1. Similarly 3^k for k>1 cannt be a common divisor. Lemma 2: If gcd(a,b)=1 and a^3 + b^3 = c^2 then either (1) a+b and a^2-a*b+b^2 are both squares or (2) a+b and a^2-a*b+b^2 are both 3 times squares. proof: a^3+b^3 = (a+b)*(a^2-a*b+b^2) = c^2 and lemma 1. Now complete parametric solutions can be given for both (1) and (2). The technique to find these is the standard method of finding all rational points on a quadratic curve when you know one point, by intersecting with lines of rational slope through the point. To get integers we need to write the slope as a quotient of integers and multiply by appropriate squares to eliminate denominators. I omit details. But the results are: (1) if a+b and a^2-a*b+b^2 are both squares then a parameterized: solution of a^3+b^3=c^2 is: a = -4*n*(-2*n+m)*(n^2-m*n+m^2) b = (m-n)*(m+n)*(7*n^2-4*m*n+m^2) c = (n^2-4*m*n+m^2)*(13*n^4-14*m*n^3+6*m^2*n^2-2*m^3*n+m^4) Note: a+b = (n^2-4*m*n+m^2)^2 and a^2-a*b+b^2 = (13*n^4-14*m*n^3+6*m^2*n^2-2*m^3*n+m^4)^2 Note 2: m and n are integers so certainly a and b cannot be prime. Note 3: it is possible in this case for a and b to be relatively prime and c not divisible by 12. (2) if a+b and a^2-a*b+b^2 are both 3 times squares then a parameterized solution for a^3+b^3=c^2 is: a = 3*m^4-n^4+6*m^2*n^2 b = -3*m^4+n^4+6*m^2*n^2 c = 6*m*n*(3*m^4+n^4) Note: a+b = 12*m^2*n^2 a^2-a*b+b^2 = 3*(3*m^4+n^4)^2, and each is 3 times a square. Note 2: if m and n have the same parity then a and b are not relatively prime, so for relatively prime a,b exactly one of m,n is even. In this case a+b is 48 times a square and c is divisible by 12. Theorem: if a,b are prime and a^3 + b^3 = c^2, then 48 divides a+b with the quotient a square, and 12 divides c. proof: since a and b are both prime we are not in case (1) above, see note 2 of that case. So we are in case (2) above and the result follows by Note 2 of that case. The parameterization in (2) allows us to very quickly find prime numbers a and b such that a^3 + b^3 = c^2. Here are the first 30, sorted by c, extending Hugo Pfoertner's data. Following Paul Leopardi's notation let d=sqrt((a+b)/48). m n a b c c/12 a+b d 1 2 11 37 228 19 48 1 6 5 8663 2137 812340 67695 10800 15 8 7 28703 8929 4935504 411292 37632 28 10 7 56999 1801 13608420 1134035 58800 35 8 11 44111 48817 14218512 1184876 92928 44 11 8 86291 6637 25354032 2112836 92928 44 10 11 87959 57241 29463060 2455255 145200 55 7 16 16931 133597 48880608 4073384 150528 56 13 14 246011 151477 135516108 11293009 397488 91 9 20 54083 334717 194057640 16171470 388800 90 16 11 367823 3889 223078944 18589912 371712 88 17 14 552011 127717 412662012 34388501 679728 119 14 23 457511 786697 763311948 63609329 1244208 161 15 22 571019 735781 764539380 63711615 1306800 165 16 25 765983 1154017 1409359200 117446600 1920000 200 23 16 1586531 38557 1998370272 166530856 1625088 184 21 26 1915163 1662229 3408412644 284034387 3577392 273 22 29 2437751 2446777 5397667572 449805631 4884528 319 15 38 16139 3882661 7650577620 637548135 3898800 285 21 34 2305883 3811669 8224333236 685361103 6117552 357 26 29 4074743 2747449 9401817516 783484793 6822192 377 25 32 3963299 3716701 10658164800 888180400 7680000 400 19 40 1296563 5634637 13456391280 1121365940 6931200 380 28 31 5440991 3600097 14413082712 1201090226 9041088 434 25 44 4683779 9836221 32471808600 2705984050 14520000 550 22 47 2238023 10591849 34633513596 2886126133 12829872 517 32 37 9682703 7139569 35661291456 2971774288 16822272 592 30 41 8681639 9473161 38787516180 3232293015 18154800 615 29 44 8142803 11395309 44940252984 3745021082 19538112 638 29 46 8321723 13032949 52820789196 4401732433 21354672 667 We can also find silly prime curiosities such as: let a,b be the 99 digit primes 516311827796740838771674147960359381895640671834628612875993558021288236937751709944319900868251071 and 164347159141325505114613844644012165100482223256876273832159859778089839875893509442524862953510657 respectively. Then a^3 + b^3 is the square of an integer and a/b = 3.14159265358979323846259.. Jim Buddenhagen

The columns headed a, b, c/12, and d are now in the
OEIS as sequences
A099806 ,
A099807 ,
A098970 ,
A099808 ,
respectively. Finally OEIS has a sequence
A099809 , for
(a^{2} - a*b + b^{2})/3 not shown in the table above.

In summary, sequences in OEIS relating to a^3 + b^3 = c^2, where a and b are prime and c is a positive integer, are related as follows: a: A099806 these primes are -1 mod 12 b: A099807 these primes are 1 mod 12 c/12: A098970 submitted by Hugo Pfoertner. sqrt((a+b)/48): A099808 sqrt((a^2-a*b+b^2)/3) A099809, so corresponding to the factorization a^3+b^3 = (a+b)*(a^2-a*b+b^2), i.e. corresponding to c^2 = (a+b)*(a^2-a*b+b^2) one has: (12*A098970)^2 = (48*A099808^2) * (3*A099809^2). The 48*3 gives 12^2, i.e. the 12 in Hugo's c/12 sequence.