Euler's Lucky numbers are positive integersnsuch thatm2 −m+nis a prime number form= 0, …,n− 1.

How do we prove that if n is lucky, then n is prime?

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- Jul 6th 2010, 04:32 PMnoviceEuler's Lucky number
Euler's Lucky numbers are positive integers

*n*such that*m*2 −*m*+*n*is a prime number for*m*= 0, …,*n*− 1.

How do we prove that if n is lucky, then n is prime? - Jul 6th 2010, 04:59 PMmelese
Try proving this by assuming to the contrary that $\displaystyle n $ is composite, or 1.

- Jul 6th 2010, 05:20 PMAlso sprach Zarathustra
Interesting post!

In my(poor) opinion to prove this we can't do what mr. melese offered...

Here is a link I think it be useful! Conjecture 17. The Ludovicus conjecture about the Euler trinomials - Jul 6th 2010, 05:20 PMchiph588@
If $\displaystyle n $ wasn't prime, then $\displaystyle n=ab $ where $\displaystyle n>a>1 $.

Now let $\displaystyle m=a $: $\displaystyle a^2-a+ab = a(a-1+b) $ which is composite. Thus $\displaystyle n $ isn't lucky. So by contraposition the claim follows. - Jul 6th 2010, 05:23 PMAlso sprach Zarathustra
- Jul 6th 2010, 05:24 PMchiph588@
- Jul 6th 2010, 05:32 PMmelese
- Jul 6th 2010, 06:07 PMchiph588@
- Jul 6th 2010, 06:38 PMmelese
But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.

- Jul 6th 2010, 06:40 PMchiph588@
- Jul 6th 2010, 06:51 PMnovice
I found in a different book that $\displaystyle 1 \leq m \leq n-1$.

It said the first 6 lucky numbers are 2, 3, 5, 11, 17 and 41 .

It should make sense to have m=0 since $\displaystyle 0^2-0+2 = 2$, which is prime. - Jul 6th 2010, 07:02 PMmelese
- Jul 6th 2010, 07:03 PMchiph588@
- Jul 6th 2010, 07:14 PMnovice
I spent half a day trying it many different ways. I attempted to prove this by contradiction at the beginning, but to prove $\displaystyle m^2-m+n \Rightarrow n$ being a composite is rather difficult because there isn't a mathematical expression for prime numbers, so I decided to prove it by contrapositive, but my main difficulty was here:

I let $\displaystyle n=ab$ where $\displaystyle 1<a<n$ and $\displaystyle 1<b<n$ and $\displaystyle a,b \in \mathbb{N}$.

Since $\displaystyle 1 \leq m \leq n-1$, I let $\displaystyle m=ab-1$.

I ended up with $\displaystyle m^2-m+n= (ab-1)^2-(ab-1)+ab= a^2b^2-2ab+2$, which led me to nowhere, and I decided to post the question.

I found your technique quite interesting. I see why you chose $\displaystyle m=a$. It makes sense since $\displaystyle 1<a<n$. That's quite creative. - Jul 8th 2010, 03:27 PMelim
To show lucky => prime, simply let m = 0

clearly prime numbers are not necessarily lucky (n=7)

Are there infinite many lucky number?