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I've searched for solutions online and I found them, but there's still one solutions that i can't understand. here there are some solutions to the problem, most of them start by showing that $k \geq 4$ is impossible, one in particular say:

Assume $k \geq 4$. Hence we get the chains $$n=a_1a_k=a_2a_{k-1}=\cdots$$$$m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=\cdots \implies u=a_1+a_k=a_2+a_{k-1}=\cdots$$Hence, the roots of the quadratic $x^2-ux+n=0$ are all the pairs $\{a_1,a_k\},\{a_2,a_{k-1}\}, \cdots$ which is a contradiction as $a_1<a_2<\cdots<a_k$.

I've read all the other solutions, but I still don't understand were this $x^2-ux+n=0$ quadratic formula came from, and why its solutions must be $\{a_1,a_k\},\{a_2,a_{k-1}\}, \cdots$.

Thanks in advance to everyone reading this.

Bill Dubuque
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Lorenzo Spada
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    "all the divisors for some $m$" is a bit unclear. Clearly $1$ is not in that list, so I guess you mean to exclude $1$. Are you also excluding $m$? – lulu Mar 13 '24 at 19:43
  • I'm assuming here that you mean: All the divisors of $n$ except $1$ and $n$. The polynomial $x^2-ux+n$ factors into both $(x-a_1)(x-a_k)$ or $(x-a_2)(x-a_{k-1})$. This is a problem because if you apply $x=a_1$, then the first factorization gives you zero, but the second one doesn't. – Tornado Mar 13 '24 at 19:59
  • Duplicate of prior-linked question - see the Lemma there. – Bill Dubuque Mar 14 '24 at 02:25
  • @Tornado the problem is that I don't understand where the polynomial X^2 - ux + n came from. I've read the answers and now I understand that it's true, but how would I have come up with that? – Lorenzo Spada Mar 14 '24 at 08:31
  • @Lorenzo That's explained by the equivalences in the Lemma in the dupe, i,e, the pairs having the same sum $s$ and product $p$ is equivalent to them being the roots of $,x^2-s,x+p.\ \ $ – Bill Dubuque Mar 14 '24 at 19:44

1 Answers1

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You have for each $i$, $$\begin{align*} m &= (a_i + 1)(a_{k+1-i}+1) \\ &= a_i a_{k+1-i} + (a_i +a_{k+1-i}) + 1 \\ &= n + (a_i +a_{k+1-i}) + 1 \end{align*}$$

Which gives you $\forall i,j \, a_i +a_{k+1-i} = a_j +a_{k+1-j} = u$

But seeing this formula $(a_i + 1)(a_{k+1-i}+1) = (-1-a_i )(-1-a_{k+1-i})$ you should see that it is a polynomial evaluated in $-1$ with roots $a_i , \,a_{k+1-i}$, which is $$p(x) = (a_i -x)(a_{k+1-i}-x) = (x-a_i )(x-a_{k+1-i}) = x^2 - ux - n$$ Written like that it is obvious that the roots are the one described in your proof. And this polynomial does not depend on $i$ hence all the pairs $\{a_i,a_{k+1-i}\}$ are in the set of its roots.

julio_es_sui_glace
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  • $(a_i + 1)(a_{k+1-i}+1) = (-1-a_i )(-1-a_{k+1-i})$ how did you get this? Also i don't understand why we are using quadratic formulas, why can we say that if $a + b = t and ab = z$ we can find them with (x-a)(x-b) = 0

    Edit: i got why we are using quadratic formula $a = t - b$ this means $b^2 - bu + z = 0$ so b and a have only one pair of answers. But from the way people are referring to this it seems that they found it some kind of obvious, there's something that I'm still not getting.

     – Lorenzo Spada Mar 14 '24 at 08:53
  • sorry for double comment. maybe what they find obvious is that if we have $a^2x + bx + c$ than $b = -(x_1 + x_2)a$ and $c = x_1 x_2*a$ ? – Lorenzo Spada Mar 14 '24 at 09:03
  • For the second one, yew, just develop, and for the first one I just multiplied each term by $(-1)$, since $(-1)^2 = 1$. You should consider working on polynomials in order to understand them better – julio_es_sui_glace Mar 16 '24 at 14:07
  • ok thanks, i understood now – Lorenzo Spada Mar 17 '24 at 11:35