(n1, n2, n3) such that b1 b2 = b3

kunny

(n1, n2, n3) such that b1 b2 = b3

**Posted:** Sat Feb 21, 2009 6:21 pm

_________________
Today's calculation of Integral Digest
Hang in there, students.

BG Yoda

$a_1a_2\left(\sum_{j = 0}^{n_1}k^j\right)\left(\sum_{j = 0}^{n_2}k^j\right) = a_3\left(\sum_{j = 0}^{n_3}k^j\right)$

$\Leftrightarrow a_1a_2(k^{n_1 + 1} - 1)(k^{n_2 + 1} - 1) = a_3(k^{n_3 + 1} - 1)(k - 1)$

We have $k^{n_i + 1} - 1\equiv - 1(\mod k^2)$
$\Rightarrow a_1a_2\equiv - a_3(k - 1)\equiv k^2 - a_3(k - 1)(\mod k^2)$

Let $a_1 = m$ , $a_1 = n$ and $a_3 = t$ . $(m,n,t\in\{1,2,\dots,k - 1\})$

Then $k^2 - (k - 1)t\equiv mn (\mod k^2)$ . Clearly, since neither $mn$ nor $k^2 - (k - 1)t$ can be greater than $k^2$ , then $mn = k^2 - (k - 1)t$ .
$k^2 - (k - 1)t\equiv 1(\mod k - 1)$ $\Rightarrow mn = s(k - 1) + 1 (1\le s\le k - 2)$ ,since $1\le mn\le (k - 1)^2$ .
$\Rightarrow (t + s)(k - 1) = (k + 1)(k - 1)$ or which is equivalent - $t = k + 1 - s$ $(\Rightarrow s\ge 2)$ .

$(s(k - 1) + 1)(k^{n_1 + 1} - 1)(k^{n_2 + 1} - 1) = (k + 1 - s)(k^{n_3 + 1} - 1)(k - 1)$

If $\min\{n_1,n_2,n_3\}\ge 2$ , then $k^3 - (k - 1)(k + 1 - s)\equiv s(k - 1) + 1(\mod k^3)$ which is impossible. Hence WLOG $n_1 = 1$ ( $n_3$ can't be 1 since then LS>RS).

$\Rightarrow (s(k - 1) + 1)(k + 1)(k^{n_2 + 1} - 1) = (k + 1 - s)(k^{n_3 + 1} - 1)$ .

Again - if $\min\{n_2,n_3\}\ge 2$ , then $k^2s + k - s + 1 = (s(k - 1) + 1)(k + 1)\equiv (k + 1 - s) (\mod k^3)$ , which is impossible. $\Rightarrow n_2 = 1$ .

$\Rightarrow (s(k - 1) + 1)(k + 1)(k^2 - 1) = (k + 1 - s)(k^{n_3 + 1} - 1)$ .
$\Leftrightarrow sk^3 + (s + 1)k^2 + (2 - s)k + 1 - s = k^{n_3 + 1} +$ $\dots + k^2 + k - sk^{n_3} - \dots - sk - s + k^{n_3} + \dots + k + 1$ . It's obvious that $n_3 > 2$ .
$\Rightarrow sk^3 + (s + 1)k^2 = k^{n_3 + 1} +$ $\dots + k^2 - sk^{n_3} - \dots - sk^2 + k^{n_3} + \dots + k^2$ .
Knowing $2\le s\le k - 2$ , we can reject with confidence the case $n_3\ge 5$ .

$1. n_3 = 4$ can be rejected, too, easily.
$2. n_3 = 3$ gives us $k^4 = 2(s - 1)k^3 + (2s - 1)k^2$

Therefore, if there existed some pair $n_1, n_2, n_3$ , then $(n_1, n_2, n_3) = (1,1,3)$ ; $a_1a_2 = s(k - 1) + 1$ , $a_3 = k + 1 - s$ and $k^2= 2(s - 1)k + (2s - 1)$ for some $2\le s\le k - 2$ . But this is impossible.

je4ko · New Member Offline Joined: 28 Oct 2007 Posts: 12

#3

Of course there is a mistake Razz

. It's in:
$- a_3(k - 1)\equiv a_3k(\mod k^2)$

**Posted:** Tue Feb 24, 2009 5:28 pm

Anavel_Gato · Poincare Conjecture Online Joined: 11 Feb 2009 Posts: 102

#4

To simplify,let $(a_1,a_2,a_3) = (a,b,c)\in \{1,2,...,k - 1\}$ and $(n_1 + 1,n_2 + 1,n_3 + 1) = (p,q,r)$ all greater than 1.
We should find possible $(p,q,r)$ such that
$ab(k^p - 1)(k^q - 1) = c(k - 1)(k^r - 1)$ (*) for some $k\ge 2$ .

since $p,q,r\ge 2$ ,we have $ab\equiv - (k - 1)c\pmod{k^2}$ ,let $ab = - (k - 1)c + uk^2$ with a positive integer $u$ ,
then $u = \frac {ab + (k - 1)c}{k^2}\le 2\left(1 - \frac {1}{k}\right)^2\Longrightarrow u = 1$ ,hence we have $ab = - (k - 1)c + k^2$ (**)
and we easily see that $ab\ge 2k - 1 > c$ since $c\le k - 1$ .

Suppose all $p,q,r > 2$ ,we have
$ab\equiv - (k - 1)c\pmod{k^3}$ by (**) we have $- (k - 1)c + k^2\equiv - (k - 1)c\pmod{k^3}\Longrightarrow k^2\equiv 0\pmod{k^3}$ which is contradiction,
hence at least one of $p,q,r$ is $2$ ,but suppose $r = 2$ then $LHS > RHS$ in (*) because $ab > c$ ,so WLOG $p = 2$ and $r > 2$ .
then (*) is equivalent to
$ab(k + 1)(k^q - 1) = c(k^r - 1)$ (***)

and suppose $q > 2$ ,we have $(k + 1)ab\equiv c\pmod{k^3}$ so let $(k + 1)ab = c + vk^3$ with a positive integer $v$ ,combined with (**) we have $(k + 1)( - (k - 1)c + k^2) = c + vk^3$ hence $v = 1 + \frac {1 - c}{k}$ so $v = 1$ and $c = 1$ ,
thus $k^2 - k + 1 = ab\le (k - 1)^2 = k^2 - 2k + 1$ ,which is contradiction,hence we have $q = 2$ .
(***) is equivalent to
$ab(k + 1)(k^2 - 1) = c(k^r - 1)$ (****)
since $c < ab\le (k - 1)^2$ we have
$(k + 1)(k^2 - 1)( = k^3 + k^2 - k - 1) < k^r - 1\le (k - 1)^2(k + 1)(k^2 - 1) (=k^5 - k^4 - 2k^3 + 2k^2 + k - 1)$ ,
and we easily see it is impossible when $r = 3$ or $r\ge 5$ ,hence we have $r = 4$ ,
then (****) becomes $ab = c(k - 1)$ which is possible successfully if we let $(a,b,c) = (k - 1,1,1)$

hence the only possible solution is $(p,q,r) = (2,2,4)$ ,that is $(n_1,n_2,n_3) = \boxed{(1,1,3)}$ .

**Posted:** Tue Feb 24, 2009 10:31 pm

BG Yoda

#5

Are you sure Anavel_Gato

when $(a_1,a_2,a_3)=(k-1,1,1)$ and $(n_1,n_2,n_3) = (1,1,3)$
then it must be $(k-1)(k^2-1)^2=(k^4-1)(k-1)$ , which is far away from the truth.

**Posted:** Wed Feb 25, 2009 12:10 am

BG Yoda

#6

**Posted:** Wed Feb 25, 2009 2:46 pm

Anavel_Gato · Poincare Conjecture Online Joined: 11 Feb 2009 Posts: 102

#7

**Posted:** Wed Feb 25, 2009 3:21 pm

BG Yoda

#8

Cool.

Very good, Gato.

**Posted:** Wed Feb 25, 2009 4:01 pm