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Number of positive integer solutions for given equation

  • Thread starter songoku
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Homework Statement


Let n be an odd integer ≥ 5. Find the number of triplets (x, y, z) of positive integers which satisfy the equation
x + y + 2z = n

Homework Equations


Do not know

The Attempt at a Solution


Let n = 2k + 1, k ≥ 2

x + y + 2z = n
2z = 2k + 1 - (x + y) ≤ 2k + 1 - 2 (because x + y ≥ 2)
2z ≤ 2k - 1
Then stuck

The answer is (n - 1)(n - 3)/4

Thanks
 

Answers and Replies

  • #2
RUber
Homework Helper
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Sometimes these are easiest to look at with a few examples.
If n = 5, then (1,2,1) and (2,1,1) are your only choices.
If n = 7, you have more. For z = 1, you have (1,4,1), (2,3,1),(3,2,1),(4,1,1) and for z=2 you have (1,2,2),(2,1,2).
What I see is that for any z, of which you are limited in choices, you will have an appropriate number of choices for (x+y)=n-2z.

So how many options for z?
For each of those options how many choices for y do you have? Once z and y and defined, you have only one choice for x.
 
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Likes songoku
  • #3
1,274
29
Sometimes these are easiest to look at with a few examples.
If n = 5, then (1,2,1) and (2,1,1) are your only choices.
If n = 7, you have more. For z = 1, you have (1,4,1), (2,3,1),(3,2,1),(4,1,1) and for z=2 you have (1,2,2),(2,1,2).
What I see is that for any z, of which you are limited in choices, you will have an appropriate number of choices for (x+y)=n-2z.

So how many options for z?
For each of those options how many choices for y do you have? Once z and y and defined, you have only one choice for x.
I think I get it. Thanks
 

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