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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

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    $\begingroup$ 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
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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.

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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 $n>10^6$.

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  • $\begingroup$ ...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

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