How to solve $5=a+2b$, $3=c+2d$ with $ad-bc = pm 1$?
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In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:
1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?
2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?
algebra-precalculus
asked Dec 19 '18 at 17:27
72D72D
417117
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add a comment |
$begingroup$
In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:
1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?
2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?
algebra-precalculus
asked Dec 19 '18 at 17:27
72D72D
417117
$endgroup$
add a comment |
$begingroup$
In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:
1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?
2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?
algebra-precalculus
asked Dec 19 '18 at 17:27
72D72D
417117
$endgroup$
In 'theory', if we choose a point $(x',y')$ on fundamental lattice it can be 'achieved' by any point $(x,y)$ if $ad-bc = pm 1$ in the system of equations $x'=ax+by$ and $y'=cx+dy$. But how in practice it is done? That is:
1- solving two equations with four unknowns $a,b,c,d in mathbb{Z}$ with one constraint $ad-bc pm 1$ doesn't give a unique matrix?
2- which integers in $5=a.1+b.2$, $3=c.1+d.2$ hold such that $ad-bc = pm 1$?
algebra-precalculus
algebra-precalculus
asked Dec 19 '18 at 17:27
72D72D
417117
asked Dec 19 '18 at 17:27
72D72D
417117
edited Dec 20 '18 at 7:45
72D
asked Dec 19 '18 at 17:27
72D72D
417117
asked Dec 19 '18 at 17:27
72D72D
417117
asked Dec 19 '18 at 17:27
72D72D
417117
417117
add a comment |
add a comment |
1 Answer
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Solve the two equations individually first.
$a+2b=5$ gives $a=2n+1, b=2-n.$
$c+2d=3$ gives $c=2m+1, d=1-n.$
Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.
EDIT
I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
$endgroup$
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
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active
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$begingroup$
Solve the two equations individually first.
$a+2b=5$ gives $a=2n+1, b=2-n.$
$c+2d=3$ gives $c=2m+1, d=1-n.$
Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.
EDIT
I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
$endgroup$
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
add a comment |
$begingroup$
Solve the two equations individually first.
$a+2b=5$ gives $a=2n+1, b=2-n.$
$c+2d=3$ gives $c=2m+1, d=1-n.$
Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.
EDIT
I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
$endgroup$
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
add a comment |
$begingroup$
Solve the two equations individually first.
$a+2b=5$ gives $a=2n+1, b=2-n.$
$c+2d=3$ gives $c=2m+1, d=1-n.$
Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.
EDIT
I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
$endgroup$
Solve the two equations individually first.
$a+2b=5$ gives $a=2n+1, b=2-n.$
$c+2d=3$ gives $c=2m+1, d=1-n.$
Now $ad-bc= 3n-5m -1,$ so you have two more linear equations to solve. In each case, you solve them in terms of a parameter $s$ and then substitute back in to the solutions above.
EDIT
I'm assuming you know how to solve a linear diophantine equation in $2$ unknowns. If you don't, make a comment, and I'll explain.
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
edited Dec 19 '18 at 18:21
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
answered Dec 19 '18 at 18:14
saulspatzsaulspatz
16k31331
16k31331
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
add a comment |
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
$begingroup$
$d=1-m$ ........
$endgroup$
– lhf
Dec 21 '18 at 9:16
add a comment |
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