Linear equations with modulus
$begingroup$
For a given prime $P$ and given coefficients $a$, $b$, $c$, $d$, $V$, $W$, I need to determine $x$ and $y$ from this system:
$ax mod P + by mod P = V$
$cx mod P + dy mod P = W$
Note that the values $V$ and $W$ can be larger than $P$.
linear-algebra
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
$endgroup$
add a comment |
$begingroup$
For a given prime $P$ and given coefficients $a$, $b$, $c$, $d$, $V$, $W$, I need to determine $x$ and $y$ from this system:
$ax mod P + by mod P = V$
$cx mod P + dy mod P = W$
Note that the values $V$ and $W$ can be larger than $P$.
linear-algebra
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
$endgroup$
1
$begingroup$
Do you mean $ax+ by equiv V (mathrm{mod} P)$ ?
$endgroup$
– mouthetics
Dec 28 '18 at 2:40
$begingroup$
there is an underlying skill: if I give you a number $n$ that is not divisible by $p,$ do you know how to find a number $t$ so that $tn equiv 1 pmod p ; ? ; ; $
$endgroup$
– Will Jagy
Dec 28 '18 at 2:40
$begingroup$
@mouthetics i believe that is what is meant, but also with the constraint that $V,W$ can be greater than $P$ (although this wouldn't change the answer).
$endgroup$
– John Doe
Dec 28 '18 at 2:41
1
$begingroup$
Does $ Pnmid ad-bc? $
$endgroup$
– Bill Dubuque
Dec 28 '18 at 2:47
add a comment |
$begingroup$
For a given prime $P$ and given coefficients $a$, $b$, $c$, $d$, $V$, $W$, I need to determine $x$ and $y$ from this system:
$ax mod P + by mod P = V$
$cx mod P + dy mod P = W$
Note that the values $V$ and $W$ can be larger than $P$.
linear-algebra
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
$endgroup$
For a given prime $P$ and given coefficients $a$, $b$, $c$, $d$, $V$, $W$, I need to determine $x$ and $y$ from this system:
$ax mod P + by mod P = V$
$cx mod P + dy mod P = W$
Note that the values $V$ and $W$ can be larger than $P$.
linear-algebra
linear-algebra
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
edited Dec 28 '18 at 10:19
Lucian Boca
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
asked Dec 28 '18 at 2:32
Lucian BocaLucian Boca
1085
1085
1
$begingroup$
Do you mean $ax+ by equiv V (mathrm{mod} P)$ ?
$endgroup$
– mouthetics
Dec 28 '18 at 2:40
$begingroup$
there is an underlying skill: if I give you a number $n$ that is not divisible by $p,$ do you know how to find a number $t$ so that $tn equiv 1 pmod p ; ? ; ; $
$endgroup$
– Will Jagy
Dec 28 '18 at 2:40
$begingroup$
@mouthetics i believe that is what is meant, but also with the constraint that $V,W$ can be greater than $P$ (although this wouldn't change the answer).
$endgroup$
– John Doe
Dec 28 '18 at 2:41
1
$begingroup$
Does $ Pnmid ad-bc? $
$endgroup$
– Bill Dubuque
Dec 28 '18 at 2:47
add a comment |
1
$begingroup$
Do you mean $ax+ by equiv V (mathrm{mod} P)$ ?
$endgroup$
– mouthetics
Dec 28 '18 at 2:40
$begingroup$
there is an underlying skill: if I give you a number $n$ that is not divisible by $p,$ do you know how to find a number $t$ so that $tn equiv 1 pmod p ; ? ; ; $
$endgroup$
– Will Jagy
Dec 28 '18 at 2:40
$begingroup$
@mouthetics i believe that is what is meant, but also with the constraint that $V,W$ can be greater than $P$ (although this wouldn't change the answer).
$endgroup$
– John Doe
Dec 28 '18 at 2:41
1
$begingroup$
Does $ Pnmid ad-bc? $
$endgroup$
– Bill Dubuque
Dec 28 '18 at 2:47
1
1
$begingroup$
Do you mean $ax+ by equiv V (mathrm{mod} P)$ ?
$endgroup$
– mouthetics
Dec 28 '18 at 2:40
$begingroup$
Do you mean $ax+ by equiv V (mathrm{mod} P)$ ?
$endgroup$
– mouthetics
Dec 28 '18 at 2:40
$begingroup$
there is an underlying skill: if I give you a number $n$ that is not divisible by $p,$ do you know how to find a number $t$ so that $tn equiv 1 pmod p ; ? ; ; $
$endgroup$
– Will Jagy
Dec 28 '18 at 2:40
$begingroup$
there is an underlying skill: if I give you a number $n$ that is not divisible by $p,$ do you know how to find a number $t$ so that $tn equiv 1 pmod p ; ? ; ; $
$endgroup$
– Will Jagy
Dec 28 '18 at 2:40
$begingroup$
@mouthetics i believe that is what is meant, but also with the constraint that $V,W$ can be greater than $P$ (although this wouldn't change the answer).
$endgroup$
– John Doe
Dec 28 '18 at 2:41
$begingroup$
@mouthetics i believe that is what is meant, but also with the constraint that $V,W$ can be greater than $P$ (although this wouldn't change the answer).
$endgroup$
– John Doe
Dec 28 '18 at 2:41
1
1
$begingroup$
Does $ Pnmid ad-bc? $
$endgroup$
– Bill Dubuque
Dec 28 '18 at 2:47
$begingroup$
Does $ Pnmid ad-bc? $
$endgroup$
– Bill Dubuque
Dec 28 '18 at 2:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To avoid the trivial cases we assume that $pnmid abcd$.
The determinant of the system is $Delta= ad - bc$.
$$Delta_x = vd-wb text{ and }Delta_y = aw - cv.$$
If $pnmid Delta$ then $$xequiv Delta^{-1}Delta_x , (mathrm{mod } p) text{ and } yequiv Delta^{-1} Delta_y, (mathrm{mod } p)$$
If $pmid Delta$ then for the system to be compatible you must have $$begin{cases}bw-dvequiv 0; (mathrm{mod } p) \ aw -cv equiv 0; (mathrm{mod } p)end{cases}$$
But by assumption this is same as $$ wequiv db^{-1} v ; (mathrm{mod } p)$$
And the system becomes equivalent to one equation $$ ax+byequiv v ;(mathrm{mod } p),$$
Which has as solution the couples $(x,-ab^{-1}x+vb^{-1})$.
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
$endgroup$
add a comment |
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1 Answer
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$begingroup$
To avoid the trivial cases we assume that $pnmid abcd$.
The determinant of the system is $Delta= ad - bc$.
$$Delta_x = vd-wb text{ and }Delta_y = aw - cv.$$
If $pnmid Delta$ then $$xequiv Delta^{-1}Delta_x , (mathrm{mod } p) text{ and } yequiv Delta^{-1} Delta_y, (mathrm{mod } p)$$
If $pmid Delta$ then for the system to be compatible you must have $$begin{cases}bw-dvequiv 0; (mathrm{mod } p) \ aw -cv equiv 0; (mathrm{mod } p)end{cases}$$
But by assumption this is same as $$ wequiv db^{-1} v ; (mathrm{mod } p)$$
And the system becomes equivalent to one equation $$ ax+byequiv v ;(mathrm{mod } p),$$
Which has as solution the couples $(x,-ab^{-1}x+vb^{-1})$.
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
$endgroup$
add a comment |
$begingroup$
To avoid the trivial cases we assume that $pnmid abcd$.
The determinant of the system is $Delta= ad - bc$.
$$Delta_x = vd-wb text{ and }Delta_y = aw - cv.$$
If $pnmid Delta$ then $$xequiv Delta^{-1}Delta_x , (mathrm{mod } p) text{ and } yequiv Delta^{-1} Delta_y, (mathrm{mod } p)$$
If $pmid Delta$ then for the system to be compatible you must have $$begin{cases}bw-dvequiv 0; (mathrm{mod } p) \ aw -cv equiv 0; (mathrm{mod } p)end{cases}$$
But by assumption this is same as $$ wequiv db^{-1} v ; (mathrm{mod } p)$$
And the system becomes equivalent to one equation $$ ax+byequiv v ;(mathrm{mod } p),$$
Which has as solution the couples $(x,-ab^{-1}x+vb^{-1})$.
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
$endgroup$
add a comment |
$begingroup$
To avoid the trivial cases we assume that $pnmid abcd$.
The determinant of the system is $Delta= ad - bc$.
$$Delta_x = vd-wb text{ and }Delta_y = aw - cv.$$
If $pnmid Delta$ then $$xequiv Delta^{-1}Delta_x , (mathrm{mod } p) text{ and } yequiv Delta^{-1} Delta_y, (mathrm{mod } p)$$
If $pmid Delta$ then for the system to be compatible you must have $$begin{cases}bw-dvequiv 0; (mathrm{mod } p) \ aw -cv equiv 0; (mathrm{mod } p)end{cases}$$
But by assumption this is same as $$ wequiv db^{-1} v ; (mathrm{mod } p)$$
And the system becomes equivalent to one equation $$ ax+byequiv v ;(mathrm{mod } p),$$
Which has as solution the couples $(x,-ab^{-1}x+vb^{-1})$.
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
$endgroup$
To avoid the trivial cases we assume that $pnmid abcd$.
The determinant of the system is $Delta= ad - bc$.
$$Delta_x = vd-wb text{ and }Delta_y = aw - cv.$$
If $pnmid Delta$ then $$xequiv Delta^{-1}Delta_x , (mathrm{mod } p) text{ and } yequiv Delta^{-1} Delta_y, (mathrm{mod } p)$$
If $pmid Delta$ then for the system to be compatible you must have $$begin{cases}bw-dvequiv 0; (mathrm{mod } p) \ aw -cv equiv 0; (mathrm{mod } p)end{cases}$$
But by assumption this is same as $$ wequiv db^{-1} v ; (mathrm{mod } p)$$
And the system becomes equivalent to one equation $$ ax+byequiv v ;(mathrm{mod } p),$$
Which has as solution the couples $(x,-ab^{-1}x+vb^{-1})$.
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
answered Dec 28 '18 at 3:27
moutheticsmouthetics
50937
50937
add a comment |
add a comment |
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1
$begingroup$
Do you mean $ax+ by equiv V (mathrm{mod} P)$ ?
$endgroup$
– mouthetics
Dec 28 '18 at 2:40
$begingroup$
there is an underlying skill: if I give you a number $n$ that is not divisible by $p,$ do you know how to find a number $t$ so that $tn equiv 1 pmod p ; ? ; ; $
$endgroup$
– Will Jagy
Dec 28 '18 at 2:40
$begingroup$
@mouthetics i believe that is what is meant, but also with the constraint that $V,W$ can be greater than $P$ (although this wouldn't change the answer).
$endgroup$
– John Doe
Dec 28 '18 at 2:41
1
$begingroup$
Does $ Pnmid ad-bc? $
$endgroup$
– Bill Dubuque
Dec 28 '18 at 2:47