# Thread: Help finding all ordered pairs satisfying?

1. ## Help finding all ordered pairs satisfying?

I don't really know where to start, I know 9000 is 2^3*3^2*5^3, and m and n have to be of different parities.

m,n
4,898
7,559
8,496
14,293
23,176
24,168
39,93
44,78
71,27
74,23

3. ## Re: Help finding all ordered pairs satisfying?

Originally Posted by Ilikebugs

I don't really know where to start, I know 9000 is 2^3*3^2*5^3, and m and n have to be of different parities.
Originally Posted by DenisB
m,n
4,898
7,559
8,496
14,293
23,176
24,168
39,93
44,78
71,27
74,23
Denis, thank you for that solution. However, don't you think if a person can program then such a solution is simple?
Question: is there a solution without using a program?
If not then what is the point of the question?

4. ## Re: Help finding all ordered pairs satisfying?

Why do you think Denis got that from a program?

Since we know the prime factorization of 9000, we can write pairs a, b such that ab= 9000 then get m= a+ 1, n= (b- m)/2.

2(4500) so m= 1, n= (4500- 1)/2 is not an integer.
3(3000) so m= 2, n= (3000- 2)/2= 2998/2= 1497. (m, n)= (2, 1497).
4(2250) so m= 3, n= (2250- 3)/2 is not an integer.
5(1800) so m= 4, n= (1800- 5)/2 is not an integer.
6(1300) so m= 5, n= (1300- 5)/2 is not an integer.
8(1125) so m= 7, n= (1125- 7)/2= 559. (m, n)= (7, 559).
etc.

5. ## Re: Help finding all ordered pairs satisfying?

Originally Posted by HallsofIvy
Why do you think Denis got that from a program?
Since we know the prime factorization of 9000, we can write pairs a, b such that ab= 9000 then get m= a+ 1, n= (b- m)/2.

2(4500) so m= 1, n= (4500- 1)/2 is not an integer.
3(3000) so m= 2, n= (3000- 2)/2= 2998/2= 1497. (m, n)= (2, 1497).
4(2250) so m= 3, n= (2250- 3)/2 is not an integer.
5(1800) so m= 4, n= (1800- 5)/2 is not an integer.
6(1300) so m= 5, n= (1300- 5)/2 is not an integer.
8(1125) so m= 7, n= (1125- 7)/2= 559. (m, n)= (7, 559).
etc.
Because it was just a listing.
You gave the reason.

6. ## Re: Help finding all ordered pairs satisfying?

with $a=m+1$ , $b=2 n+m$ , and $a b = 9000$

Case $a$ even, $b$ must be odd since $a+b$ is odd

now $9000=2^3 3^2 5^3$ so $a=8 c$ where $c$ is a divisor of $3^2 5^3$

$b>a$ implies $\frac{9000}{8c}>8c$ which implies $1\leq c\leq 11$ and therefore $c=1,3,5,9$

so in this case we have four solutions

Case $a$ odd

7. ## Re: Help finding all ordered pairs satisfying?

Or turn into a quadratic:

m = {-2n - 1 +- sqrt[(2n + 1)^2 - 8n + 36000]} / 2

...so [(2n + 1)^2 - 8n + 360000] = a square

8. ## Re: Help finding all ordered pairs satisfying?

Originally Posted by Idea
with $a=m+1$ , $b=2 n+m$ , and $a b = 9000$

Case $a$ even, $b$ must be odd since $a+b$ is odd

now $9000=2^3 3^2 5^3$ so $a=8 c$ where $c$ is a divisor of $3^2 5^3$

$b>a$ implies $\frac{9000}{8c}>8c$ which implies $1\leq c\leq 11$ and therefore $c=1,3,5,9$

so in this case we have four solutions

Case $a$ odd
So, DenisB missed:
2, 1499

9. ## Re: Help finding all ordered pairs satisfying?

Originally Posted by SlipEternal
So, DenisB missed:
2, 1499
Yes. I did have it at top of list, then went to add (m,n) headings and
somehow erased the 2,1499...I'll go stand in the corner for 14.99 minutes...