A positive integer $a$ is called a double number if it has an even number of digits (in base 10) and its base 10 representation has the form $a = a_1a_2 \cdots a_k a_1 a_2 \cdots a_k$ with $0 \le a_i \le 9$ for $1 \le i \le k$, and $a_1 \ne 0$. For example, $283283$ is a double number. Determine whether or not there are infinitely many double numbers $a$ such that $a + 1$ is a square and $a + 1$ is not a power of $10$.
Problem
Source: Indian IMOTC 2013, Team Selection Test 4, Problem 1
Tags: modular arithmetic, Diophantine equation, number theory proposed, number theory
mathdebam
03.08.2013 17:37
Basically,we have to prove whether the equation $n(10^k+1)=s^2-1$ have infinitely many positive integer solutions with $10^{k-1}\le n \le 10^k$ except the trivial solution $n=10^k-1,s=10^k$.
mathdebam
04.08.2013 18:59
And a pell's equation $s^2-(10^k+1)x^2=1$ where the conventional $d=10^k+1$ and $n=x^2$.Now the problem looks something tackalable with $10^{k-1}\le n<10^k.$
dionescu
05.08.2013 08:37
How would you know that the Pell equation has a solution satisfying the given constraint? Also, I could not find any example for small k (< 8) where a perfect square solution $n$ exists. In general, it seems that for most $n$ there are several solutions that satisfy $s-n = $ constant, but I could not come up with a general form ...
dibyo_99
23.03.2014 17:39
Once we know that there are infinitely many solutions (I was told so by the attendees of the camp), its not very difficult to construct the solutions.
$n= \frac{9}{11} \left( \frac{9(10^{22p+1}+1)}{11}+2 \right)$
ThinkingSuppose that $n(10^k+1)=t^2-1$ with $10^{k-1} \le n < 10^k$. First, take $k$ odd, say $k=2m+1$. Obviously, $t < 10^k$. Since $11|10^{2m+1}+1$, we may take take $t-1= \frac{s(10^{2m+1}+1)}{11}$ for some $s \in (1,11)$. Now, we need \[ n(10^{2m+1}+1) = \frac{s(10^{2m+1}+1)}{11}\left( \frac{s(10^{2m+1}+1)}{11}+2 \right) \]\[ \implies 121n= s(s(10^{2m+1}+1)+22)\]Therefore, $121|s(s(10^{2m+1}+1)+22)$. Since $s \le 11$, so $(s,11)=1$. As a result, $121|s(10^{2m+1}+1)+22$. Therefore, $s(10^{2m+1}+1) \equiv 99 \pmod{121}$. Take $s=9$. Then, we need \[10^{2m+1}+1 \equiv 11 \pmod{121} \implies \frac{10^{2m+1}+1}{10+1} \equiv 1 \pmod{11}\]\[\implies 10^{2m}-10^{2m-1}+\cdots +1 \equiv 1 \pmod{11} \implies 2m \equiv 0 \pmod{11}\]So, it is sufficient to have $11|m$. And therefore, we have generated infinitely many solutions.
P.S. USA people would probably hate me for creating a solution with an infinitude of 9/11's
TheBernuli
03.04.2014 14:30
Maybe I don't understand. Why we can't take only the numbers in the form $a=10^{10k-2}-1$?
dibyo_99
03.04.2014 16:58
If you do that, then $a+1$ is a power of $10$.
TheBernuli
03.04.2014 18:16
I thought that $a+1$ couldn't be in form $k^{10}$, not $10^k$. Sorry -.-
Goutham
03.04.2014 18:26
This problem was pretty simple but almost everyone lost 1 mark because we didnt prove the fact that a+1 is not a power of 10, even if it was pretty obvious in most constructions, silly mistake indeed