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## Hexagonal Triangular Squares

Is there a 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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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/… – David Speyer Jul 14 2010 at 22:36

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.

(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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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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Since 2 people already made the same mistake, I suggest to insert "centered" in the question. – Ale De Luca Jul 14 2010 at 23:07

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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