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$6$ and $210$ share the property that both are the products of both two and three consecutive numbers. $6$ is $2\times3$ and $1\times2\times3$ and $210$ is $14\times15$ and $5\times6\times7$. It was easy enough to write a program to search for more numbers with this property, I found that there were no more up to at least $1{,}000{,}000{,}000{,}000$ ($1$ Trillion). But it is beyond me to prove that there are either no more numbers like this or to find the next one. Any ideas?

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If $(b-1)b(b+1)=a(a+1),$ solving for $a,$ we get $$\frac{-1\pm \sqrt{4b^3-4b+1}}2$$ So, we need $4b^3-4b+1$ to be perfect square as it is always odd – lab bhattacharjee May 29 '13 at 16:04
There might or might not be an elementary proof of non-existence of other solutions available, but in general, this looks like finding rational points on elliptic curve. The following thread might be useful. – Peter Košinár May 29 '13 at 16:07
Magma says the only integer points on $b^3-b = a^2-a$ have $b \in \{-1,0,1,2,6\}$, confirming that $6$ and $210$ are the only non-zero solutions. – Erick Wong May 29 '13 at 16:09
@PatrickDaSilva, I'm not sure either. Just disclosed that if it helps. – lab bhattacharjee May 29 '13 at 16:16
By looking up, I find that this is elliptoc curve "37a1", with Mordell-Weil group generated by $(0,0)$. At least that simplifies the process of looking for all rational solutions, but I still didn't stumble upon other integer points. – Hagen von Eitzen May 29 '13 at 16:30
up vote 24 down vote accepted

In his book Diophantine equation,(page $257-258$) L.J.Mordell proved that the equation $$y(y+1)=x(x+1)(x+2)$$ has only the integer solutions $x=-1,-2,0,1,5.$ enter image description here

enter image description here

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If Mordell says so... +1 – Patrick Da Silva May 29 '13 at 17:58

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