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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?

Hagen von Eitzen
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User1
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  • I'll be glad to assist. just explain the question again please :) – DanielY May 29 '13 at 16:01
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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
  • 6 and 210 are the only numbers less than 1 trillion that are the products of two and also three consecutive numbers. I'm searching for either proof that there are no more (a good guess since there are no more numbers like 6 and 210 for a long time) or for a way to find more numbers like this. :) – User1 May 29 '13 at 16:05
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    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
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    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
  • What is "Magma"? – User1 May 29 '13 at 16:11
  • @lab bhattacharjee : What you are saying is that you are looking for integer solutions to the elliptic curve $y^2 = 4b^3−4b+1$. To be honest I didn't study elliptic curves in a while so I don't know if this would be an optimal way to solve this problem. – Patrick Da Silva May 29 '13 at 16:14
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    @PatrickDaSilva, I'm not sure either. Just disclosed that if it helps. – lab bhattacharjee May 29 '13 at 16:16
  • @ErickWong : There is an issue with 'computer solutions' ; we don't know if their algorithm behind the scenes only finds the 'small' solutions or if they are actually right. How do you expect Magma to find a solution beyond its internal capacity? OP says he tested above 1 trillion. – Patrick Da Silva May 29 '13 at 16:16
  • @user1 : After lab bhattacharjee's comment, I think it is beyond all of us. Maybe an elliptic curve theorist could come around and solve this in the integers? – Patrick Da Silva May 29 '13 at 16:17
  • @PatrickDaSilva I agree with the theoretical risk of computer solutions (notice I did not post it as an answer), but you are aware that Magma is not just testing brute force solutions but actually has tables of elliptic curves? This is Magma's specialty (not that it is necessarily perfect), and 1 trillion is extremely small for a computer algebra system's capacity. – Erick Wong May 29 '13 at 16:27
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    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

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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

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lsr314
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