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This question already has an answer here:

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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marked as duplicate by Jack D'Aurizio, user147263, Mike Pierce, Paramanand Singh, Jonas Meyer Jul 3 '15 at 5:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Related: When is $\frac{n(n+1)(n+2)}6$ a perfect cube? $\endgroup$ – punctured dusk Jul 1 '15 at 12:54
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    $\begingroup$ @barto : looks similar , but really related ?!? $\endgroup$ – user228168 Jul 1 '15 at 12:56
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    $\begingroup$ I 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. $\endgroup$ – Maciek Jul 1 '15 at 12:57
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    $\begingroup$ @Maciek : Just to make it more non-elegant ; I think $n=48$ also works ... $\endgroup$ – user228168 Jul 1 '15 at 13:03
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    $\begingroup$ The answer to this question (math.stackexchange.com/questions/778233/…) tells us that there are no solutions other than $n=1,2,48$. Since your equation is $\binom{n+2}{3}=m^2$. $\endgroup$ – Pjotr5 Jul 1 '15 at 13:09
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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 $n\in\{1,2,48\}$.

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