Try this beautiful problem from Algebra based on Largest Common Divisor .
For natural numbers
But try the problem first...
Answer:$3$
PRMO-2018, Problem 21
Pre College Mathematics
First hint
At first we have to find out the divisors that satisfy the equation \( xy=x+y+(x,y)\) .so we assume that \(x\)=ak and \(y\)=bk and try to find out the divisors
Can you now finish the problem ....
Second Hint
Let \(x\) =ak and \(y\)=bk, then (x,y)=k and (a,b)=1
Therefore \(xy=x+y+(x,y)\)
\(\Rightarrow abk^2=ka+kb+k\)
\(\Rightarrow kab=a+b+1\)
Can you finish the problem...
Final Step
\(k=1\Rightarrow ab=a+b+1\Rightarrow a=1=\frac{2}{b-1}\)
For \(a\in \mathbb N\), then (b-1) divides 2
\(\Rightarrow b-1=1,2\)
\(\Rightarrow b=2,3\)
Therefore a=1+2=3 and a=1+1=2
\(\Rightarrow (x,y)=(3,2) or (2,3)\)
Now for \(x=y\Rightarrow(x,y)=x\)
so \(xy=x+y(x<y)\)
\(\Rightarrow x^2=3x\)
\(\Rightarrow x^2 -3x=0\)
\(\Rightarrow x(x-3)=0\)
\(\Rightarrow x=3 or 0\)
Therefore \(x\in \mathbb N\Rightarrow x=3,y=3\)
Therefore \((x,y)=(3,3)\)
Hence total number of pairs =3
Try this beautiful problem from Algebra based on Largest Common Divisor .
For natural numbers
But try the problem first...
Answer:$3$
PRMO-2018, Problem 21
Pre College Mathematics
First hint
At first we have to find out the divisors that satisfy the equation \( xy=x+y+(x,y)\) .so we assume that \(x\)=ak and \(y\)=bk and try to find out the divisors
Can you now finish the problem ....
Second Hint
Let \(x\) =ak and \(y\)=bk, then (x,y)=k and (a,b)=1
Therefore \(xy=x+y+(x,y)\)
\(\Rightarrow abk^2=ka+kb+k\)
\(\Rightarrow kab=a+b+1\)
Can you finish the problem...
Final Step
\(k=1\Rightarrow ab=a+b+1\Rightarrow a=1=\frac{2}{b-1}\)
For \(a\in \mathbb N\), then (b-1) divides 2
\(\Rightarrow b-1=1,2\)
\(\Rightarrow b=2,3\)
Therefore a=1+2=3 and a=1+1=2
\(\Rightarrow (x,y)=(3,2) or (2,3)\)
Now for \(x=y\Rightarrow(x,y)=x\)
so \(xy=x+y(x<y)\)
\(\Rightarrow x^2=3x\)
\(\Rightarrow x^2 -3x=0\)
\(\Rightarrow x(x-3)=0\)
\(\Rightarrow x=3 or 0\)
Therefore \(x\in \mathbb N\Rightarrow x=3,y=3\)
Therefore \((x,y)=(3,3)\)
Hence total number of pairs =3