Talk:Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2

In Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2/Mistake/First Edition:

An integer is the sum of a sequence of consecutive integers if and only if it is not a power of $2$.

The property stated holds only for positive integers.

For example, $−16$ cannot be so represented, but it is not a power of $2$ as such.

This mistake has been corrected in David Wells: Curious and Interesting Numbers (2nd ed.).

However the case where $2 a + n - 1 = \pm 1$ can be abused to give a representation:

$-16 = -16 + \paren {-15} + ... + 14 + 15$.

Through this exploit every integer can be represented as a sum of consecutive integers (trivially), including the powers of $2$:

$x = x + \paren {-|x - 1| + \paren {-|x - 2|} ... + |x - 2| + |x - 1|}$

$4 = 4 + 3 + 2 + 1 + 0 + \paren {-1} + \paren {-2} + \paren {-3}$

There doesn't seem to be any way to preserve the significance of the theorem without sacrificing its validity, e.g. on the odd primes.

Perhaps "sum of less than n consecutive integers"? RandomUndergrad (talk) 06:30, 19 April 2020 (EDT)

I wonder if it should say:

"Then $n$ can be expressed as the summation of $2$ or more consecutive (strictly) positive integers if and only if $n$ is not a power of $2$." That seems to work, e.g.:

$3 = 1 + 2$, $5 = 2 + 3$, $6 = 1 + 2 + 3$, $7 = 3 + 4$

(Odd integers are of course simple: $2 n + 1 = n + \paren {n + 1}$.)

What do you think? Then I will publish another page in the errata list for the 2nd edition, to complement the one in the 1st edition.

Hint: if you include an equals sign in a parameter to a template, enclose it in double braces, so the Mediawiki software does interpret as a label-parameter pair itself:

See: Help:FAQ/Technical questions/Can't get a maintenance template to register my parameter

Your work is excellent, and greatly appreciated. I'm getting old so my abilities (never exceptional) are fading, so I can't come up with proofs any more. All I can do is review the work of others. :-) --prime mover (talk) 07:36, 19 April 2020 (EDT)

If $\dfrac {m \paren {2 a + m - 1} } 2 = 1$ then $2 = m \paren {2 a + m - 1}$ and so $m = 2$ (as $m \ge 2$ by hypothesis) leading to $2 a + m - 1 = 1$ and so after algebra $a = 0$.

Hence $1 = 0 + 1$ but $a$ is not a strictly positive integer.

Does that work for you? --prime mover (talk) 07:43, 19 April 2020 (EDT)

It still occurs to be that the sufficient condition needs to be adjusted to restrict the terms to being strictly positive.

I think we just have to show that $\dfrac n d - m$ is positive.

In fact the proof given doesn't work for, say, $n = 7$. --prime mover (talk) 07:46, 19 April 2020 (EDT)

It seems to work; I'll grind out the details. My intuition is to divide by $2$ until you can't anymore.

By the way, is the name of the page wrong? There is no divisibility anywhere. RandomUndergrad (talk) 10:53, 19 April 2020 (EDT)

You're right, I don't know where that came from. I will fix it in due course.

Right, I think this is all done. Good job. --prime mover (talk) 18:43, 19 April 2020 (EDT)