## BdMO National Secondary/Higher Secondary 2018/6

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### BdMO National Secondary/Higher Secondary 2018/6

Find all integer solution ($m,n$) for the following equation:
$3(m^2+n^2)-7(m+n)=-4$

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO National Higher Secondary 2018#6

Hint

Abdullah Al Tanzim
Posts: 22
Joined: Tue Apr 11, 2017 12:03 am

### Re: BdMO National Higher Secondary 2018#6

$3(m^2+n^2)- 7(m+n) =-4$
Substitute $(m+n)=x$ and $(m^2+n^2)=y$
So, $7x-3y=4$ and the general solution of this diophantine equation is $x=3k+1$, $y=7k+1$ where $k$ is an integer.
So, $(m+n)=3k+1$............$(1)$
$(m^2+n^2)=7k+1$...........$(2)$
From $(2)$ we get that
$m^2+n^2=7k+1$
or,$[(m+n)^2 +(m-n)^2]/2=7k+1$
or,$(m-n)^2= 14k+2-(3k+1)^2$
or, $(m-n)^2=8k+1-9k^2$
As $(m-n)^2\geq 0$,
$9k^2 \leq 8k+1$
From this inequality where $k$ is an integer the possible values for $k$ are $0,1$
Case $1$: when $k=0$
$m^2+n^2=1$ and $m+n=1$
The integer solutions of these two equations are $(0,1)$ and $(1,0)$
Case $2$: when $k=1$
$m^2+n^2=8$ and $m+n=4$
The only integer solution of these two equations is $(2,2)$
So, the solutions are $(1,0), (0,1), (2,2)$
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

### Re: BdMO National Secondary/Higher Secondary 2018/6

$3(m^2+n^2)-7(m+n)=-4$
$\Rightarrow 6m^2+6n^2-14m-14n=-8$
$\Rightarrow 36m^2+36n^2-84m-84n=-48$
$\Rightarrow (6m)^2-84m+7^2+(6n)^2-84n+7^2=-48+49+49$
$\Rightarrow (6m-7)^2+(6n-7)^2=50$

$\blacktriangleright$ $(6m-7,6n-7)\in\{(\pm 1,\pm 7),(\pm 7,\pm 1),(\pm 5,\pm 5)\}$
$\blacktriangleright$ $\boxed{(m,n)\in\{(1,0),(0,1),(2,2)\}}$ Last edited by nahin munkar on Wed Jan 09, 2019 5:09 pm, edited 1 time in total.
Reason: A typo.