Square pyramidal numbers and Tetrahedral numbers

[Helios]

There are square pyramidal numbers and tetrahedral numbers, defined

Square pyramidal numbers = n ( n + 1 )( 2 n + 1) / 6
1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, ...

Tetrahedral numbers = n ( n + 1 )( n + 2 ) / 6

1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, ...

and I was wondering if there's a number(s) besides 1 that is both.

 

[hamster143]

I have to guess no. Purely for probabilistic reasons. For any number n, the probability that there is a pyramidal number equal to n(n+1)(n+2)/6 is ~1/n^2, falling off too fast as n -> infinity. So, for example, the probability of a hit for n>100 is ~0.01. Once we've checked the first 100 n's, we can be fairly sure that there won't be any hits beyond that.

But I have no idea how to give a proper proof.

 

[csco]

The answer is no. You are looking for solutions of the equation:

<br /> \frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6}<br />

Cancelling terms we get

<br /> 2n+1 = n+2<br />

Solving for n you find n = 1 as the only solution

 

[CRGreathouse]

The answer is no. You are looking for solutions of the equation:

<br /> \frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6}<br />

Cancelling terms we get

<br /> 2n+1 = n+2<br />

Solving for n you find n = 1 as the only solution

The OP is looking for solutions to

<br /> \frac {n(n+1)(2n+1)}{6} = \frac {m(m+1)(m+2)}{6}<br />
not

<br /> \frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6}<br />
.

 

[hamster143]

According to Mathworld, 1 is the only solution, and this fact was only proven in 1988 (so, no easy proof is forthcoming).

 

[ramsey2879]

According to Mathworld, 1 is the only solution, and this fact was only proven in 1988 (so, no easy proof is forthcoming).

really? Can you give a cite?

 

[CRGreathouse]

MathWorld gives

Beukers, F. "On Oranges and Integral Points on Certain Plane Cubic Curves." Nieuw Arch. Wisk. 6, 203-210, 1988.​

but Nieuw Archief voor Wiskunde's online archives only go back to 2000.