BdMO National Secondary 2015/2

[samiul_samin]

How many pairs of integers ($m,n$) satisfy the equation $m+n=mn$ ?

[samiul_samin]

Hint

1.Multiplication is nothing but long addition
2.Two conseqative numbers are always Co-prime

Answer

$\fbox 0$

[samiul_samin]

Solution

$m+n=mn$
$\Rightarrow m+n=m+m+m+m+m+m+...+m $ (n times)
$\Rightarrow n=m+m+m+m+m....+m+m$ {(n-1)times}
$\Rightarrow n=m(n-1)\Rightarrow \dfrac {n}{n-1}=m$
But $n$ and $n-1$ are co-prime.
So, $m$ cannot be an integer.
A Contrudiction
So,there are no such $(m,n) $that satisfies the equation $m+n=mn$

[mac0220]

The solution is ( m,n ) = (2,2)
n and n-1 are co prime unless n-1=1
which makes n=m=2

[samiul_samin]

The solution is ( m,n ) = (2,2)
n and n-1 are co prime unless n-1=1
which makes n=m=2

Yes,you are right.It was a very foolosh mistake of me.Sorry.

[Ohin01]

Solution

$m+n=mn$
$\Rightarrow m+n=m+m+m+m+m+m+...+m $ (n times)
$\Rightarrow n=m+m+m+m+m....+m+m$ {(n-1)times}
$\Rightarrow n=m(n-1)\Rightarrow \dfrac {n}{n-1}=m$
But $n$ and $n-1$ are co-prime.
So, $m$ cannot be an integer.
A Contrudiction
So,there are no such $(m,n) $that satisfies the equation $m+n=mn$

(0,0) is also a solution
LHS=0/(0-1)=0/-1=0
RHS=0×0=0