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Number Theory Diophantine equation

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I got problem with this exemple (It's 'Exempel 1' and yes it's on Swedish), I follow my book method and get,
\(\displaystyle x=169-35K, \ y= -468-97K\)
I get the 'homogen soloution' that \(\displaystyle x=-35k, \ y=97k\)
'Solve equation \(\displaystyle 97x+35y=13\)'

Regards,
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
Hello MHB,
I got problem with this exemple (It's 'Exemple 1' and yes it's on Swedish), I follow my book method and get,
\(\displaystyle x=169-35K, \ y= -468-97K\)
I get the 'homogen soloution' that \(\displaystyle x=-35k, \ y=97k\)
'Solve equation \(\displaystyle 97x+35y=13\)'

Regards,
Erm... what is your question?
 

Petrus

Well-known member
Feb 21, 2013
739
Solve equation \(\displaystyle 97x+35y=13\)
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,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
Solve equation \(\displaystyle 97x+35y=13\)
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?)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

Petrus

Well-known member
Feb 21, 2013
739
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
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?
 

Petrus

Well-known member
Feb 21, 2013
739
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?
\(\displaystyle x=169-35K, \ y= -468-97K\)
cause that website says the answer shall be
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
so this answer is correct?
\(\displaystyle x=169-35K, \ y= -468-97K\)
cause that website says the answer shall be
The website answer is correct. If you add a multiple of 35 to $x$ then you have to subtract that multiple of 97 from $y$.
 

Petrus

Well-known member
Feb 21, 2013
739
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 \(\displaystyle x=x_0 , y=t_0\) (particulate solution) that is what we find we euclidmens algorithmen.
*Find all soloution to homogeneous equation \(\displaystyle ax'+by'=0\) and that is the one i do it wrong then...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
*Find all soloution to homogeneous equation \(\displaystyle ax'+by'=0\) 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...
 

Petrus

Well-known member
Feb 21, 2013
739
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 \(\displaystyle x'=-kb, \ y'= ka\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
that is the part I strugle, my book say \(\displaystyle x'=-kb, \ y'= ka\)
That works as well.
Can you substitute that solution in the equation?
 

Petrus

Well-known member
Feb 21, 2013
739

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908

Petrus

Well-known member
Feb 21, 2013
739

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
\(\displaystyle 35(-35k)+97*97k=0\) ?
I kind of lost track of the equation. :)
But the equation was $97x'+35y'=0$.

Can you redo the substitution with this equation?
 

Petrus

Well-known member
Feb 21, 2013
739
I kind of lost track of the equation. :)
But the equation was $97x'+35y'=0$.

Can you redo the substitution with this equation?
\(\displaystyle 97(-35k)+35*97k\)
my book says \(\displaystyle x=x_0+x' , y=y_0+y'\)
and our \(\displaystyle x_0=169\) and \(\displaystyle y_0=-468\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
\(\displaystyle 97(-35k)+35*97k\)
Do you see that this is zero?

my book says \(\displaystyle x=x_0+x' , y=y_0+y'\)
and our \(\displaystyle x_0=169\) and \(\displaystyle y_0=-468\)
Yes?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
\(\displaystyle 97(-35k)+35*97k\)
my book says \(\displaystyle x=x_0+x' , y=y_0+y'\)
and our \(\displaystyle x_0=169\) and \(\displaystyle y_0=-468\)
That means the answer is \(\displaystyle x=169-35K, \ y= -468-97K\)
Erm.

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

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

See the difference?
 

Petrus

Well-known member
Feb 21, 2013
739
Erm.

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

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

See the difference?
Finally, I did find my mistake...! I somehow did have \(\displaystyle -97k\) in my paper... and confused myself..! Can you confirm that this is correct answer?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
Finally, I did find my mistake...! I somehow did have \(\displaystyle -97k\) 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...
 

Petrus

Well-known member
Feb 21, 2013
739
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 alot better! Hey look at the bright side, you gained some 'thanks' ;)

Regards,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
Yep. Thanks! ;)