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

Question

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

(A) 1

(B) 2

(C) 3

(D) 4

(E) more than 4

nick1816 · Answer

mn-m=n
m(n-1)=n
m=n/n-1
m=1+1/n-1
m will be integer when 1/n-1 is an integer
1/n-1 is an integer when n= 0 or 2
when n=0, m=0 
and when n=2, then m=2

There are 2 pairs possible.

Vinit800HBS · Answer

The given equation can be written as:

mn - m = n
m*(n-1) = n
m = n/(n-1) = 1 + 1/(n-1)

Now, we have to look for factors of 1. There are two factors: 1 and -1

So, n-1 = 1 or n-1 = -1
n = 2, m = 2
n = 0, m = 0

Thus, only 2 pairs are possible. 
Option 2 is our answer choice

Posted from my mobile device

Kinshook · Answer

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

mn - m = n
m(n-1) = n
m = n/(n-1)

If n=0; m = 0
If n =1; Not feasible
If n=2; m=2

n & (n-1) are co-prime n/(n-1) is an integers only when (n-1) is 1 or -1

Only 2 solutions are possible

IMO B

Bankut · Answer

Identify possible pairing of odd/even to help.

odd+odd = even
 even+even=even 
odd+even= odd

odd*odd=odd
odd*even=even
 even*even=even

Since m+n = mn ---> meaning (m = even , n = even) are the possible outcome.

then, input several paring until get an answer.

mn - m = n
m(n-1) = n

Alec0 · Answer

For anyone wondering how to go from step 3 to step 4..

1 =   \frac{(n-1)}{(n-1)}

\frac{n}{(n-1)} = \frac{(n - 1 + 1)}{(n-1)} = \frac{(n-1)}{(n-1)}+ \frac{1}{(n-1)} = 1 + \frac{1}{(n-1)}

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