# On $\sum_{k=1}^nk^3 = x^3 + y^3$ with $x,y \ge 1$

My question is related to https://oeis.org/A269839.

It is well-known that there are parametric families of solutions for cubes that are sums of consecutive cubes: https://arxiv.org/pdf/1603.08901.pdf. Famous and simplest one is $$6^3 = 3^3 + 4^3 + 5^3$$ which Euler noted.

I have also curiosity about solutions to $$\sum_{k=1}^nk^3 = x^3 + y^3$$ with $$x,y \ge 1$$, although it seems that there is no any reference on it. Largest solution that I found as below. $$\sum_{k=1}^{524^{3}}k^3 = 107131073934081017703266616960000 = 32201140654^3 + 41934379346^3$$

In particular, I am not sure about that there are additional probably interesting values of $$n$$ such as $$7^{2}$$, $$61^{3}$$, $$293^{3}$$, $$440^{3}$$ and $$524^{3}$$.

Question. Can be infinitely many solutions to diophantine equation $$\sum_{k=1}^nk^3 = x^3 + y^3$$ with $$x,y \ge 1$$ ?

Any comment is welcome in order to suggest helpful ideas on question. Also any additional term is very welcome.

Thanks.

• I think you are wolcome to study solutions of $2(n^4+n^2)+4n^3= 8(x^3 + y^3)$ – zeraoulia rafik Aug 6 at 13:02
• @zeraouliarafik: Yes, we can rewrite the equation based on $\sum_{k=1}^nk^3 = T_{n}^{2}$ as you did, $n^2 + 2n^3 + n^4 = 4(x^3+y^3)$. Best regards. – Alkan Aug 6 at 16:52
• @Alkan: $\sum_{k=1}^nk^3 = x^3 + y^3$ can be transformed to $v^2 = -3s^4+(3n^4+6n^3+3n^2)s$. We get $(n,s,x,y)=(2112278, 237469827, 70918018, 166551809).$ – Tomita Aug 7 at 0:15
• @Tomita: Many thanks for your comment. Yes, I agree with your transformation. Is this larger solution that you found yet? May you look further ranges and can you confirm $291^{3}$ ? I am also not sure that there are other solution(s) between $2112278$ and $291^{3}$. Best regards. – Alkan Aug 7 at 4:44
• @Tomita: Thanks, I understand. Your example is also confirmation of last term of OEIS entry. Best regards. – Alkan Aug 7 at 5:20