Thread: Diophantine equation

Question:

Find all solutions that are positive integers of the Diophantine equation:

25x + 35y = 380

I need to check my answer. Not sure If I have the correct solutions

x > 0
y > 0

25x + 35 = 380 ==> x = 13.8

25 + 35y = 380 ==> y = 10.14

0 < x < 14
0 < y < 11

25x + 35y = 380 ==> 5x + 7y = 76
25x + 175k + 105 = 380 making
x = 11-7k


25(11) + 35(4) = 275 + 105 = 380
25(4) + 35(8) = 100 + 280 = 380

Solution
K(0,1)
X(11,4)
Y (3,8)


Not sure if I expressed my answer corretly.

Originally Posted by math_cali

Question:

Find all solutions that are positive integers of the Diophantine equation:

25x + 35y = 380

Now, this is actually the way you can get the answer to the other problem you posted without guessing.
---
There is a cool way to do these using continued fractions but I do not think you number theory course does those until the end of the semester if not at all.

We need to find,

Apply Euclidean algorithm,



Now work backwards,




Thus,

Multiply by 76,

Thus,

oh ok thanks alot I thought I was suppost to be looking for exact numbers. Now I understand and see how I could apply this in the previous problem I posted

Hello, math_cali!

Find all solutions that are positive integers of the Diophantine equation:
. .

Your answers are correct!
I would write them as: .

I kind of understand how I got the solutions when I wrote it out the frist time, but how would you apply the solutions to what ThePerfectHacker showed me when he worked it out?

Hello, math_cali!

Here's a really primitive approach . . .

Find all positive integer solutions of: .

Divide by 5: .

Solve for

This can be written: .

. . Hence: . [1]


Since is an integer, must some integer
. . We have: .

This can be written: .

. . Hence: . [2]


Since is an integer, then must be some integer

. . We have: .

. . Hence: . [3]


Since is an integer, must be even: .

Substitute into [3]: .

Substitute into [2]: .

Substitute into [1]: .


We have found all solutions: . for any integer


Since and must be positive integers:
. .

And there are only two solutions: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Of course, this is a long and ponderous method . . . I warned you.

There are far better methods (such as the Euclidean Algorithm).

I wanted to show you that, if desperate,
. . you can baby-talk your way through these problems.

..
We want positive solutions.

Thus,


Solve,


Thus,

We need integers.
Those are,

Put them into the x,y equations to get the two possible solutions.