# Euler's Lucky number

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• Jul 6th 2010, 05:32 PM
novice
Euler's Lucky number
Euler's Lucky numbers are positive integers n such that m2 − 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, 05:59 PM
melese
Try proving this by assuming to the contrary that $n$ is composite, or 1.
• Jul 6th 2010, 06:20 PM
Also 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, 06:20 PM
chiph588@
Quote:

Originally Posted by Also sprach Zarathustra
In my(poor) opinion to prove this we can't do what mr. melese offered..

If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.
• Jul 6th 2010, 06:23 PM
Also sprach Zarathustra
Quote:

Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky by contraposition.

How you can choose m like that?
• Jul 6th 2010, 06:24 PM
chiph588@
Quote:

Originally Posted by Also sprach Zarathustra
How you can choose m like that?

$m$ takes on all values less than $n$.
• Jul 6th 2010, 06:32 PM
melese
Quote:

Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.

For completness, when you start with $n$ not a prime you should also examine $n=1$(not a prime). Then what happens is that $m=0$ and $m^2-m+n=1$, and also $n$ isn't lucky.
• Jul 6th 2010, 07:07 PM
chiph588@
Quote:

Originally Posted by melese
For completness, when you start with $n$ not a prime you should also examine $n=1$(not a prime). Then what happens is that $m=0$ and $m^2-m+n=1$, and also $n$ isn't lucky.

To avoid this special case you could just assume $n$ is composite and let $m=0$.
• Jul 6th 2010, 07:38 PM
melese
But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.
• Jul 6th 2010, 07:40 PM
chiph588@
Quote:

Originally Posted by melese
But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.

But my way above takes care of 1 just like you did.
• Jul 6th 2010, 07:51 PM
novice
I found in a different book that $1 \leq m \leq n-1$.
It said the first 6 lucky numbers are 2, 3, 5, 11, 17 and 41 .
It should make sense to have m=0 since $0^2-0+2 = 2$, which is prime.
• Jul 6th 2010, 08:02 PM
melese
Quote:

Originally Posted by chiph588@
But my way above takes care of 1 just like you did.

I assume you mean:
Quote:

Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.

I don't see how the case n=1 is covered, because $n=ab$ and $a>1$.
• Jul 6th 2010, 08:03 PM
chiph588@
Quote:

Originally Posted by melese
I assume you mean:

I don't see how the case n=1 is covered, because $n=ab$ and $a>1$.

No I offered another solution later.
• Jul 6th 2010, 08:14 PM
novice
Quote:

Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.

I spent half a day trying it many different ways. I attempted to prove this by contradiction at the beginning, but to prove $m^2-m+n \Rightarrow n$ 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 $n=ab$ where $1 and $1 and $a,b \in \mathbb{N}$.
Since $1 \leq m \leq n-1$, I let $m=ab-1$.

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

I found your technique quite interesting. I see why you chose $m=a$. It makes sense since $1. That's quite creative.
• Jul 8th 2010, 04:27 PM
elim
To show lucky => prime, simply let m = 0

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

Are there infinite many lucky number?
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