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Integer Divisibility Dilemma (Posted on 2014-04-10) Difficulty: 2 of 5
Find all positive integers X and Y such that:
  • X divides Y+5, and:
  • Y divides X+3
Prove that there are no others.

See The Solution Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

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Solution Full solution with proof of completeness | Comment 5 of 6 |
From the 1st equation, x <= y + 5
From the 2nd equation, y <= x + 3

In other words, y - 3 <= x <= y + 5

Let's consider the only 9 possible cases:

a) x = y + 5
   Then (substituting in equation 2), y divides y + 8
   y can only be 1,2,4,8
   This leads to 4 solutions: (6,1), (7,2), (9,4) and (13,8)

b) x = y + 4
   Then (substituting in equation 2), y divides y + 7
   y can only be 1,7
   this gives rise to no solutions

c) x = y + 3
   Then (substituting in equation 2), y divides y + 6
   y can only be 1,2,3,6
   this gives rise to no solutions

d) x = y + 2
   Then (substituting in equation 2), y divides y + 5
   y can only be 1,5
   this gives rise (3,1)

e) x = y + 1
   Then (substituting in equation 2), y divides y + 4
   y can only be 1,2,4
   this gives rise to (2,1)

f) x = y 
   Then (substituting in equation 2), y divides y + 3
   y can only be 1,3
   this gives rise to (1,1)

g) x = y - 1
   Then (substituting in equation 2), y divides y + 2
   y can only be 1,2
   this gives rise to (1,2)

h) x = y - 2
   Then (substituting in equation 2), y divides y + 1
   y can only be 1
   this gives rise to no solutions

h) x = y - 3
   Then (substituting in equation 2), y divides y 
   But, y = x + 3
   Then (substituting in equation 1), x divides x + 8
   x can only be 1,2,4,8
   This leads to 4 solutions: (1,4), (2,5), (4,7) and (8,11)
  
This method is exhaustive, so there are no other solutions  

  Posted by Steve Herman on 2014-04-10 16:10:23
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