Hexagonal Triangular Squares - MathOverflow

Hexagonal Triangular Squares

2010-07-14T22:06:40Z

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?

Answer by Ale De Luca for Hexagonal Triangular Squares

2010-07-14T23:06:08Z

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

Answer by Max Alekseyev for Hexagonal Triangular Squares

2010-07-14T23:07:13Z

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 $n=1$, $m=1$, $p=0$.

Answer by dke for Hexagonal Triangular Squares

2010-07-14T23:08:57Z

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.)