Question

Find the number of ordered pairs (m,n) of integers that satisfy
mn = 2m + 4n.

Answer 1

Answer:

  There are 8 such ordered pairs.

Step-by-step explanation:

Notice that n divides both mn and 4n, so from mn = 2m + 4n, it follows that n also divides 2m.  That is:

• 2m = bn for some integer b.     ...(*)

Multiplying the given equation by 2 gives

• 2mn = 2×2m + 8n

Substituting (*) gives

• bn² = 2bn + 8n

Supposing for now that n≠0 (need to remember to deal with this later, if necessary), divide through by n to get

• bn = 2b + 8   ⇒   n = 2 + 8/b

So b must be a factor if 8 for n to be an integer.  That is, b must be -8, -4, -2, -1, 1, 2, 4 or 8.  Checking these one by one:

• b = -8  ⇒  n = 2-1 = 1  and  m = bn/2 = -4
• b = -4  ⇒  n = 2-2 = 0  and  m = bn/2 = 0    [the case n=0 is addressed!]
• b = -2  ⇒  n = 2-4 = -2  and  m = bn/2 = 2
• b = -1  ⇒  n = 2-8 = -6  and  m = bn/2 = 3
• b = 1  ⇒  n = 2+8 = 10  and  m = bn/2 = 5
• b = 2  ⇒  n = 2+4 = 6  and  m = bn/2 = 6
• b = 4  ⇒  n = 2+2 = 4  and  m = bn/2 = 8
• b = 8  ⇒  n = 2+1 = 3  and  m = bn/2 = 12

Putting these into the original equation confirms that they are all solutions.

So there are 8 ordered pairs in total and they are:

     (-4, 1),  (0, 0),  (2, -2),  (3, -6),  (5, 10),  (6, 6),  (8, 4),  (12, 3)

Hope this helps!