# Math Help - Euler's Lucky number

1. ## 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?

2. Try proving this by assuming to the contrary that $n$ is composite, or 1.

3. 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

4. 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.

5. 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?

6. Originally Posted by Also sprach Zarathustra
How you can choose m like that?
$m$ takes on all values less than $n$.

7. 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.

8. 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 $m=0$.

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

10. 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.

11. 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.

12. Originally Posted by chiph588@
But my way above takes care of 1 just like you did.
I assume you mean:
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 $a>1$.

13. 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.

14. 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.

15. 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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