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I've been working on the Collatz conjecture and stumbled across what I think is a major property of prime numbers: every odd number can be expressed as a sum of consecutive odd numbers except the odd primes.

Is this a new discovery?

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    And what about $1$? – 2017-01-18
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    Sums of how many consecutive odd integers? (Adding two odds gives an even, so evidently there's an odd number of terms.) – 2017-01-18
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    I'm not clear what your question is exactly. No I didn't notice that. Did that answer your question ;) – 2017-01-18
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    well this is obvious that no of consecutive odd numbers are odd. – 2017-01-18
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    oh yeah except odd number 1 @MarkBennet – 2017-01-18
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    i didn't knew that fact so thats why i just shared.@BradThomas – 2017-01-18
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    I really don't think the question should be downvoted. It may not be phrased very well, but it's coming from someone discovering something for themselves, and wondering if it's known. I've taken the liberty to tone down what I suspect is causing the "friction." OP, if I've put any words into your mouth, feel free to modify the post :) – 2017-01-18
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    thanks for modifying and also for the motivational support.:) @pjs36 – 2017-01-18
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    Also see this question and its answers: http://math.stackexchange.com/questions/299553/what-numbers-can-be-written-as-a-sum-of-two-or-more-consecutive-odd-numbers – 2017-01-18

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This is an interesting observation - to answer your question, yes, it's well known.

First I will try to clarify what I think you mean.

You can always express an odd number as a sum of one consecutive odd number - itself. I assume this is a trivial case you want to ignore, so your "consecutive" means "more than one".

Then you are right. Primes (and $1$ itself) are the only odd numbers that can't be expressed as a sum of two or more consecutive odd numbers.

You can put together a proof this way:

The sum of the first $n$ odd integers is $n^2$. That means the sum of consecutive odd integers is a difference of squares $m^2-n^2 = (m-n)(m+n)$. Then given a composite odd number $ab$ with $a$ and $b$ odd and greater than $1$ you solve $a = m+n$ and $b = m-n$ to find $m$ and $n$.

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    yeah you got me. now i understood that it is just another form of expressing product of two numbers. @Etha – 2017-01-18
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Any sum of consecutive odd numbers can also be expressed as a product of the average (which will be an integer) and the count of values. Therefore if we confine the odd numbers to positive values and the count to be $\ge 2$, the result cannot be a prime number.

For example, $$7+9+11+13+15 = \overset{\text{average}}{11}\times\overset{\text{count}}5 = 55$$

If we allow negative odd numbers into the sum, then we can arrange for the average to be $1$ and any odd number is possible. We can also make any number divisible by $4$, but not even numbers which are twice an odd number.

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    thanks for the information. – 2017-01-18