Solving Diophantine Equation: 97x+35y=13

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The forum discussion centers on solving the Diophantine equation 97x + 35y = 13 using Euclid's algorithm. The user initially derives a particular solution of x = 169 - 35K and y = -468 - 97K, but seeks clarification on the correct approach to find all solutions. Through community guidance, it is confirmed that the general solution can be expressed as x = 169 - 35K and y = -468 + 97K, with the importance of correctly applying the homogeneous solution noted. The discussion concludes with the user gaining a clearer understanding of Diophantine equations.

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Petrus
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Hello MHB,
I got problem with this http://staff.www.ltu.se/~larserik/applmath/chap10/part4.html (It's 'Exempel 1' and yes it's on Swedish), I follow my book method and get,
$$x=169-35K, \ y= -468-97K$$
I get the 'homogen soloution' that $$x=-35k, \ y=97k$$
'Solve equation $$97x+35y=13$$'

Regards,
 
Last edited:
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Petrus said:
Hello MHB,
I got problem with this http://staff.www.ltu.se/~larserik/applmath/chap10/part4.html (It's 'Exemple 1' and yes it's on Swedish), I follow my book method and get,
$$x=169-35K, \ y= -468-97K$$
I get the 'homogen soloution' that $$x=-35k, \ y=97k$$
'Solve equation $$97x+35y=13$$'

Regards,

Erm... what is your question?
 
Solve equation $$97x+35y=13$$
ehmm. I am suposed to find one soultion with gcd, eucliders algoritmen then find all other soloution, cause it got infinity soloution, for x and y. Do it make sense now?

Regards,
 
Petrus said:
Solve equation $$97x+35y=13$$
ehmm. I am suposed to find one soultion with gcd, eucliders algoritmen then find all other soloution, cause it got infinity soloution, for x and y. Do it make sense now?

Regards,

Ah, okay.
So do you know how to apply Euclid's algorithm?
(Btw, isn't this university-level stuff instead of pre-algebra?)
 
I like Serena said:
...
(Btw, isn't this university-level stuff instead of pre-algebra?)

Good call...moved to the Number Theory sub-forum.
 
I like Serena said:
Ah, okay.
So do you know how to apply Euclid's algorithm?
(Btw, isn't this university-level stuff instead of pre-algebra?)
Yes, I get x=169 and y=-468
 
Petrus said:
Yes, I get x=169 and y=-468

That looks about right.
Wolfram says x=29, y=-80, which is effectively the same answer.
(Since you can add a multiple of 35 to x if you also subtract the same multiple of 97 from y.)
So you're all good?
 
I like Serena said:
That looks about right.
Wolfram says x=29, y=-80, which is effectively the same answer.
(Since you can add a multiple of 35 to x if you also subtract the same multiple of 97 from y.)
So you're all good?
so this answer is correct?
$$x=169-35K, \ y= -468-97K$$
cause that website says the answer shall be
http://staff.www.ltu.se/~larserik/applmath/chap10/eq409.gif
 
Petrus said:
so this answer is correct?
$$x=169-35K, \ y= -468-97K$$
cause that website says the answer shall be
http://staff.www.ltu.se/~larserik/applmath/chap10/eq409.gif
The website answer is correct. If you add a multiple of 35 to $x$ then you have to subtract that multiple of 97 from $y$.
 
  • #10
Opalg said:
The website answer is correct. If you add a multiple of 35 to $x$ then you have to subtract that multiple of 97 from $y$.
I am kinda confused cause I follow the method from my book.
* find a soloution $$x=x_0 , y=t_0$$ (particulate solution) that is what we find we euclidmens algorithmen.
*Find all soloution to homogeneous equation $$ax'+by'=0$$ and that is the one i do it wrong then...
 
  • #11
Petrus said:
*Find all soloution to homogeneous equation $$ax'+by'=0$$ and that is the one i do it wrong then...

The solution is $x'=kb,\ y'=-ka$, for integers k (note the minus sign).
Substitute to verify...
 
  • #12
I like Serena said:
The solution is $x'=kb,\ y'=-ka$, for integers k (note the minus sign).
Substitute to verify...
that is the part I strugle, my book say $$x'=-kb, \ y'= ka$$
 
  • #13
Petrus said:
that is the part I strugle, my book say $$x'=-kb, \ y'= ka$$

That works as well.
Can you substitute that solution in the equation?
 
  • #14
I like Serena said:
That works as well.
Can you substitute that solution in the equation?
$$x' = -35k , \ y'= 97k$$
 
  • #15
Petrus said:
$$x' = -35k , \ y'= 97k$$

Yes.
Can you substitute that in the equation $ax′+by′=0$?
 
  • #16
I like Serena said:
Yes.
Can you substitute that in the equation $ax′+by′=0$?
$$35(-35k)+97*97k=0$$ ?
 
  • #17
Petrus said:
$$35(-35k)+97*97k=0$$ ?

I kind of lost track of the equation. :)
But the equation was $97x'+35y'=0$.

Can you redo the substitution with this equation?
 
  • #18
I like Serena said:
I kind of lost track of the equation. :)
But the equation was $97x'+35y'=0$.

Can you redo the substitution with this equation?
$$97(-35k)+35*97k$$
my book says $$x=x_0+x' , y=y_0+y'$$
and our $$x_0=169$$ and $$y_0=-468$$
 
  • #19
Petrus said:
$$97(-35k)+35*97k$$

Do you see that this is zero?

my book says $$x=x_0+x' , y=y_0+y'$$
and our $$x_0=169$$ and $$y_0=-468$$

Yes?
 
  • #20
I like Serena said:
Do you see that this is zero?
Yes,
I like Serena said:
Yes?
That means the answer is $$x=169-35K, \ y= -468-97K$$
 
  • #21
Petrus said:
$$97(-35k)+35*97k$$
my book says $$x=x_0+x' , y=y_0+y'$$
and our $$x_0=169$$ and $$y_0=-468$$

Petrus said:
That means the answer is $$x=169-35K, \ y= -468-97K$$

Erm.

You have $x'=-35k,\ y'=97k$ and $$x_0=169$$, $$y_0=-468$$

If I substitute that into $$x=x_0+x' , y=y_0+y'$$, I get
$$\qquad x=169+(-35k) , y=-468+(97k)$$.

See the difference?
 
  • #22
I like Serena said:
Erm.

You have $x'=-35k,\ y'=97k$ and $$x_0=169$$, $$y_0=-468$$

If I substitute that into $$x=x_0+x' , y=y_0+y'$$, I get
$$\qquad x=169+(-35k) , y=-468+(97k)$$.

See the difference?
Finally, I did find my mistake...! I somehow did have $$-97k$$ in my paper... and confused myself..! Can you confirm that this is correct answer?
 
  • #23
Petrus said:
Finally, I did find my mistake...! I somehow did have $$-97k$$ in my paper... and confused myself..! Can you confirm that this is correct answer?

Erm, yes. It is the correct answer.
It's what the website solution said...
 
  • #24
I like Serena said:
Erm, yes. It is the correct answer.
It's what the website solution said...
Hello I like Serena,
I am sorry I just got confused, Thanks for taking your time and now I understand diophantine equation a lot better! Hey look at the bright side, you gained some 'thanks' ;)

Regards,
 
  • #25
Yep. Thanks! ;)
 

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