What is the smallest perfect square that ends in $9009$?
Problem
Source:
Tags: Diophantine Equations
mszew
16.01.2008 17:17
$ k^2=10000 n + 9009$ for $ k,n \in \mathbb N$
Taking mod $ 10$
$ k^2\equiv 9 (10)$ then $ k\equiv 3(10)$ or $ k\equiv 7(10)$
$ k=10m+r$ with $ r \in \{3,7\}$
Taking mod $ 100$
$ k^2\equiv 9 (100)$
$ (10m+3)^2 \equiv 9(100)$ then $ 60m \equiv 0 (100)$
then $ m\equiv 0 (5)$
$ (10m+7)^2 \equiv 9(100)$ then $ 140m+40 \equiv 0 (100)$
then $ m\equiv 4 (5)$
$ k=100o+r$ with $ r \in \{3,53,47,97\}$
Taking mod $ 10000$
$ k^2\equiv 9009 (10000)$
$ (100o+3)^2 \equiv 9009(10000)$
$ 600o \equiv 9000 (10000)$
$ 3o \equiv 45 (50)$
$ o \equiv 15 (50)$
$ (100o+53)^2 \equiv 9009(10000)$
$ 600o+2809 \equiv 9009 (10000)$
$ 600o \equiv 6200 (10000)$
$ 3o \equiv 31 (50)$
$ o \equiv 27 (50)$
$ (100o+47)^2 \equiv 9009(10000)$
$ 9400o+2209 \equiv 9009 (10000)$
$ 9400o \equiv 6800 (10000)$
$ 47o \equiv 34 (50)$
$ o \equiv 22 (50)$
$ (100o+97)^2 \equiv 9009(10000)$
$ 9400o+9409 \equiv 9009 (10000)$
$ 9400o+400 \equiv 0 (10000)$
$ 47o+2 \equiv 0 (50)$
$ o \equiv 34 (50)$
then minimum $ k=1503$
H.HAFEZI2000
16.05.2020 04:40
my solution is the same, $x^2 \equiv 9 \mod 1000 \implies x=t(500)\pm 3$ so $x^2 = (500t \pm 3)^2 \equiv 9009 \mod 10000$ which implies $\pm t \equiv 3 \mod 10$ now because we want the smallest solution we shall put $t=3$ and $x=500t+3=1503$