## anonymous one year ago someone could help me here , please .

1. anonymous

2. anonymous

i don't know what method i have to use there. :l

3. UsukiDoll

I only have a slight idea for the first one given n is an odd integer $\geq 5$ so your n must be an odd number greater than or equal to 5

4. UsukiDoll

x, y - positive integers.. so we just have to pick all positive numbers

5. UsukiDoll

@baad1994 try letting x = 1 , y = 3, and n = 7 for x+2y=n

6. perl

*

7. UsukiDoll

@perl I have no idea if I had the right idea for the first one.... T__________________T I think I did, but it's only some combinations

8. dumbcow

since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n - 2y , so 2y <= n-1 since x must be positive also y is pos integer 1,2,3 ... n-1 then num of possible y-values and num of pairs for any given n is (n-1)/2 total num of pairs is then just the infinite sum $\sum_{i=2}^{\infty} \frac{n-1}{2}, n = 2i +1$ which simplifies to $\sum_{i=2}^{\infty} i$

9. perl

$$\color{blue}{\text{Originally Posted by}}$$ @dumbcow since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n - 2y , so 2y <= n-1 since x must be positive also y is pos integer 1,2,3 ... n-1 $$\color{blue}{\text{End of Quote}}$$ Question. Should that say y is a pos. integer 1,2,3,.. (n-1)/2

10. dumbcow

ahh yes, thank you for the correction

11. perl

Nice solution :)