Let $\overline{abcd}$ be one of the 9999 numbers $0001, 0002, 0003, \ldots, 9998, 9999$. Let $\overline{abcd}$ be an special number if $ab-cd$ and $ab+cd$ are perfect squares, $ab-cd$ divides $ab+cd$ and also $ab+cd$ divides $abcd$. For example 2016 is special. Find all the $\overline{abcd}$ special numbers. Note: If $\overline{abcd}=0206$, then $ab=02$ and $cd=06$.
Problem
Source: Cono Sur Olympiad 2016, problem 1
Tags: number theory, Perfect Squares, cono sur
RagvaloD
18.08.2017 11:37
Let $ab-cd=x^2,ab+cd=x^2y^2,y>1$ then $2ab=x^2(y^2+1)$ $\overline{ab}+\overline{cd}|100\overline{ab}+\overline{cd} \to \overline{ab}+\overline{cd}|99\overline{ab} \to x^2y^2|99x^2(y^2+1) \to y^2|99 \to y=3$ $ab=5x^2,cd=4x^2$ and $ab<100$ so $x<5$ Answer: $0504,2016,4536,8064$
ugesaurio
30.07.2018 18:12
RagvaloD wrote: Let $ab-cd=x^2,ab+cd=x^2y^2,y>1$ then $2ab=x^2(y^2+1)$ $\overline{ab}+\overline{cd}|100\overline{ab}+\overline{cd} \to \overline{ab}+\overline{cd}|99\overline{ab} \to x^2y^2|99x^2(y^2+1) \to y^2|99 \to y=3$ $ab=5x^2,cd=4x^2$ and $ab<100$ so $x<5$ Answer: $0504,2016,4536,8064$ I think you don't need $y>0$, meaning it is possible to have $cd = 00$. You're missing the trivial ones: $0100, 0400, 0900, 1600, 2500, 3600, 4900, 6400$ and $8100$.
Golub_Srecko
30.07.2018 19:46
ab-cd|ab+cd ab-cd=x² ab+cd=x² ab+cd|100ab+cd=>ab+cd|99ab=>ab-cd|99ab(1) But ab-cd|ab+cd+(ab-cd)=2ab (2) From (1) and (2) ab-cd|ab Now ab=t•x² Therefore cd=(t-1)x² y²=(2t-1)x² But from (1) 99tx²=m•(2t-1)x² As m is an integer 2t-1|99t From here we get that possible t is 1,2,5,6,17,99 where only t=1 and t=5 have solutions. abcd={0100,0400,0900,1600,2500,3600,4900,6400,8100, 0504,2016,4536,8064}