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According to MathWorld,

Pentagonal Triangular Number: A number which is simultaneously a pentagonal number $P_n$ and triangular number $T_m$. Such numbers exist when $$\frac{1}{2}n(3n-1)=\frac{1}{2}m(m+1).$$ Completing the square gives $$(6n-1)^2-3(2m+1)^2=-2.$$ Substituting $x=6n-1$ and $y=2m+1$ gives the Pell-like quadratic Diophantine equation $$x^2-3y^2=-2,$$ which has solutions $(x,y)=(5,3),(19,11),(71,41),(265,153), \ldots$.

However, it does not state how these solutions for $(x,y)$ were obtained.

I know that the solution $(5,3)$ can be obtained by observing that $1$ is both a pentagonal and a triangular number.

Does obtaining the other solutions simply involve trial-and-error? Or is there a way to obtain these solutions?

6 Answers 6