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Squares that are equal to the sum of 3 consecutive cubes.
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%I #26 Sep 08 2022 08:45:24

%S 0,9,36,41616

%N Squares that are equal to the sum of 3 consecutive cubes.

%C m^3+(m+1)^3+(m+2)^3=3(1+m)*(3+2*m+m^2). Corresponding values of m are -1,0,1,23.

%C The equation s^2 = 3c^3 + 6c can be transformed using the substitution X = 3c, Y = 3s into Y^2 = X^3 + 18X, a form of the Weierstrass equation of an elliptic curve: Y^2 = X^3 + aX^2 + bX + c, with a = c = 0. We can now use the Sage program to show that there are no other integer solutions. - _Jaap Spies_, May 27 2007

%C Confirmed by MAGMA - see code below. - _Warut Roonguthai_, May 28 2007

%C That there are no other integer solutions is a theorem of Uchiyama, rediscovered by Cassels. For n consecutive cubes summing to a square, see A218979. - _Jonathan Sondow_, Apr 03 2014

%H J. W. S. Cassels, <a href="http://dx.doi.org/10.1017/S0017089500006030">A Diophantine equation</a>, Glasgow Math. J., 27 (1985), 11-18.

%H Saburo Uchiyama, <a href="http://projecteuclid.org/euclid.pja/1195517117">On a Diophantine equation</a>, Proc. Japan Acad., Ser. A 55 (1979), 367-369.

%t Select[Total/@Partition[Range[-2,200]^3,3,1],IntegerQ[Sqrt[#]]&] (* _Harvey P. Dale_, Aug 08 2015 *)

%o (Magma) IntegralPoints(EllipticCurve([18,0]));

%Y Cf. A027602, A218979.

%K fini,nonn,full

%O 1,2

%A _Zak Seidov_, Apr 14 2007