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March 13th, 2018, 01:15 PM   #1
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Is There A Triangular Number Proof For This?

It seems like triangular numbers are multiples of 4 if and only if the term number is 1 less than a multiple of 8 (such as 28 as the 7th term) or a multiple of 8 (such as 36 as the 8th term). Is there a proof of it?
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March 13th, 2018, 03:29 PM   #2
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There is a closed form formula for the triangular numbers given by:
\[T_n = \frac{n(n+1)}{2} \]
It's obvious that if $n = 0 \mod 8$ or $n = 7 \mod 8$, then $T_n$ is a multiple of 4. Do you see why?

On the other hand, suppose $T_n$ is a multiple of 4, then the denominator implies that $n(n+1)$ must be a multiple of 8. Therefore, there are at least 3 factors of 2 floating between $n$ and $n+1$. If all 3 factors belong to either term, then you get the converse of the above claim. However, does this always happen? Can you prove that they can't "split" the factors of 2?

Hint: $n,n+1$ are consecutive integers, so one is even and one is odd.
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Last edited by skipjack; March 13th, 2018 at 04:05 PM.
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March 14th, 2018, 05:29 AM   #3
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I know the first part. If n = 7 mod 8, n + 1 = 0 mod 8. In both cases the numerator will be a multiple of 8 and dividing by 2 makes a multiple of 4. I don't know the second part.
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