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Is there triangular number which is also a power of 2 number? $(n^2+n)/2 = 2^n$ beside 1.

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closed as off-topic by Thursday, Claude Leibovici, Tunk-Fey, Hakim, Moron Aug 3 at 10:19

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You seem to have two different $n$'s in there! –  TonyK Mar 3 at 13:09
1  
Since this is a homework question,you should also add your own thoughts and ideas to the body of this question.That will prevent people from telling you things which you already know.Also,please don't use 'homework' as a stand-alone tag.Please reconsider editing your post to add more tags and adding your own thoughts. –  rah4927 Mar 3 at 13:09
    
And no,you cannot have a triangular number equal to a power of 2. –  rah4927 Mar 3 at 13:12

3 Answers 3

up vote 1 down vote accepted

$$\frac{n(n+1)}2=\frac n2(n+1)=n\frac{n+1}2$$

Thus, either $\;n\;$ or $\;n+1\;$ is odd and thus there's a prime factor different from two...

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Hint: from your expression above, show that every triangular number has a prime factor which is not 2.

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No. Triangular numbers are of the form $\displaystyle \frac{n(n+1)}{2}$, and at least one of the two ($n$ or $n+1$) has to be an odd number. And unless $n = 0$ or $n = 1$, the odd number will prevent the triangular number from being a power of $2$.

When $n = 0$, $\displaystyle \frac{n(n+1)}{2} = 0$, which by default is not a power of any number other than itself. When $n = 1$, $\displaystyle \frac{n(n+1)}{2} = 1$, which is the trivial example (and one you are not looking for).

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