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?
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
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.
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.
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.