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Is there a centered-hexagonal, triangular, square (apart from 0 and 1)?
In other words, is there a positive integer that is simultaneously

(1) a perfect square, $n^2$, $n \ge 2$,

(2) a triangular number, $\frac{m(m+1)}{2}$, $m$ an integer,

and (3) a centered-hexagonal number, $(p+1)^3 - p^3$, $p$ an integer?

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    $\begingroup$ I'll repeat the comment I've made on this problem before: "I’m way to lazy to work out the details right now, but: the equations you are dealing with are a pair of quadratic equations in three variables. That means they probably cut out an elliptic curve. Assuming this is correct, we know from Siegel’s Theorem that there are only finitely many integral points." topologicalmusings.wordpress.com/2008/09/28/… $\endgroup$ Jul 14, 2010 at 22:36
  • $\begingroup$ Well, that combination did not work. But we can basically force a nontrivial solution for centered polygonal numbers with other sides counts for the centered polygon. Centered octagonal numbers are one place to look. These have the form $4n^2+4n+1$ which is $(2n+1)^2$. So for that case we simply have all odd triangular-square numbers belonging to the list ($1,1225,...$). Another possibility is centered enneagonal* numbers (* — Yes, I prefer the Greek word for nine like other polygons). These have the formula $(9n^2+9n+2)/2$. It turns out that $9n^2+9n+2=(3n+1)(3n+2)$, so these numbers are inva $\endgroup$ Jun 3, 2023 at 18:42
  • $\begingroup$ @StefanKohl I'm curious. If my answer had to be converted to a comment how could it fit given the 600-character limit? $\endgroup$ Jun 4, 2023 at 0:16
  • $\begingroup$ @Oscar, it didn't fit, it was cut off mid-sentence. $\endgroup$ Jun 4, 2023 at 0:22
  • $\begingroup$ Another possibility is centered enneagonal* numbers (* — Yes, I prefer the Greek word for nine like other polygons). These have the formula $(9n^2+9n+2)/2$. It turns out that $9n^2+9n+2=(3n+1)(3n+2)$, so these numbers are invariably triangular. Triangular squares that are not multiples of $3$ are included, which turn out to be the same as the odd triangular squares from the centered octagonals. Thus again $1225$ is the first nontrivial solution. $\endgroup$ Jun 4, 2023 at 0:59

3 Answers 3

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The only solution is 1 - this was a question asked in a book by Gardner and proved by Charles Grinstead, On a Method of Solving a Class of Diophantine Equations , Mathematics of Computation, Vol. 32, No. 143 (Jul., 1978), pp. 936-940.

See http://www.jstor.org/pss/2006498.

(N.B. The hexagonal numbers you are using are actually the centred hexagonal numbers - it's obvious usual hexagonal numbers are always triangular.)

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The question is equivalent to the system of quadratic Diophantine equations: $$8 n^2 = m'^2 - 1$$ $$4n^2 = 3p'^2 + 1$$ where $m'=2m+1$ and $p'=2p+1$. How to solve such systems is described in my paper: http://arxiv.org/abs/1002.1679 (see Theorem 6)

It is easy to obtain that the only nonegative solution is $n=1$, $m=1$, $p=0$.

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1225 is not a centered hexagonal number (which is the usual term for (3), see http://en.wikipedia.org/wiki/Centered_hexagonal_number )

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    $\begingroup$ Since 2 people already made the same mistake, I suggest to insert "centered" in the question. $\endgroup$ Jul 14, 2010 at 23:07

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