How many positive integers $n$ are there such that $\dfrac {n(n+1)(n+2)}6$ is a perfect square ? I know $n=1 , 2$ works ; are there any more ? Are there only finitely many such $n$ ?
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1Related: When is $\frac{n(n+1)(n+2)}6$ a perfect cube? – Bart Michels Jul 01 '15 at 12:54
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1@barto : looks similar , but really related ?!? – Jul 01 '15 at 12:56
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1I think that it's just the two. Notice that n$ (n+1) and (n+2) can have no common factors apart from 2. So all prime factors of 3 or higher need all be grouped into a single term of the product. I can finish it with case bashing, but it's not particularly elegant. – Maciek Jul 01 '15 at 12:57
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2@Maciek : Just to make it more non-elegant ; I think $n=48$ also works ... – Jul 01 '15 at 13:03
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3The answer to this question (http://math.stackexchange.com/questions/778233/is-there-any-perfect-squares-that-are-also-binomial-coefficients) tells us that there are no solutions other than $n=1,2,48$. Since your equation is $\binom{n+2}{3}=m^2$. – Pjotr5 Jul 01 '15 at 13:09
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@Pjotr5: This is a complete answer; if you post it as an answer, the question can be marked as answered. – joriki Jul 01 '15 at 13:13
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This is related to the sum of the first $k$ squares being a square. Suppose $\frac 16 k(k+1)(2k+1)=r^2$ then, with $n=2k$ we have $\frac 16 n(n+2)(n+1)=4r^2=m^2$ - and the only non-trivial solution to that is $k=24$. – Mark Bennet Jul 01 '15 at 13:15
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Your equation is the same as finding integer solutions to the equation $$ \binom{n+2}{3}=m^2. $$
The top answer to this question tells us that there are only solutions for $n\in\{1,2,48\}$.
