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

In Positive Integer is Divisible by Sum of Consecutive Integers iff not Power of 2/Mistake:


 * 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 $n$ 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. --prime mover (talk) 07:46, 19 April 2020 (EDT)