My question is: is it possible to find pentagonal numbers which are also tetrahedral? A pentagonal number is obtained by the formula: $$P_k=\frac{1}{2}k(3k-1)$$ The equivalent formula for the tetrahedral number $T_n$ is: $$T_n=\frac{1}{6}n(n+1)(n+2)$$ So the problem is to find a $T_n=P_k$ that means to solve: $$n(n+1)(n+2)=3k(3k-1)$$ with $k\in\mathbb{N}$ and $n\in\mathbb{N}$ Can someone give me a hint how to solve this equation? Thanks

  • 3
    $\begingroup$ Diophantine equations of the form quadratic in $y$ equals cubic in $x$ are generally difficult to solve, but there are some (advanced) techniques around. The keyphrase is "elliptic curve". $\endgroup$ – Gerry Myerson Jun 13 '13 at 9:20

Let $Y=3k-1,X=n+1$, then, as @GerryMyerson said in the comments, you have an elliptic curve $Y^2+Y=X^3-X$, and as such it can only have finitely many integer solutions.

You can use Sage to find them:

sage: E=EllipticCurve([0,0,1,-1,0])
sage: E
Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field
sage: E.integral_points()
[(-1 : 0 : 1), (0 : 0 : 1), (1 : 0 : 1), (2 : 2 : 1), (6 : 14 : 1)]

Or you can look it up in the elliptic curve database at LMFDB.

Three of the solutions have $Y=k=0$, the other two give $k=1,P_k=1$ and $k=5,P_k=35$, so the solutions @MarioCarneiro found are all of them.


I don't have a general solution for you, but a brute force search yielded only the 3 solutions

  • $n=k=0$ with $T_n=P_k=0$,
  • $n=k=1$ with $T_n=P_k=1$, and
  • $n=k=5$ with $T_n=P_k=35$.

Any other solution would need $n>10^6$.

  • $\begingroup$ ...with the respective pentagonal/tetrahedral numbers in question being $1$ and $35$ (also $0$ if you count that) $\endgroup$ – Sp3000 Jun 13 '13 at 10:06
  • $\begingroup$ @Sp3000 fixed${}$. $\endgroup$ – Mario Carneiro Jun 13 '13 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.