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Unknown value in the modular numerator
The Next CEO of Stack OverflowModular simple equationWhat value minimizes the error from a set of values under modular arithmetic?modular arithmetic with exponentsFind the least positive integer with remainders 1,2, and 3 when divided by 7,8, and 9 respectively.Modular Arithmetic - Finding the smallest possible length of the room in inchesModular arithmetic : Subtraction and SimplificationDetermining an unknown parameter in an equation with modular arithmeticsSystem of modular equations with unknown modulusHow to find the least integer in modular arithmetic?How to get a perfect square with this condition?
$begingroup$
I'm looking for a mathematical solution to find unknown value in the numerator as follows:
IF: $$b_1 + b_2 = 151$$
$$n = 86167$$
$$ (231 * 336 + b_1)^2 mod 86167 = 151^2 $$
$$ (25 * 336 + b_2)^2 mod 86167 = 151^2 $$
So, How can I find $b_1$ and $b_2$ mathematically ??
modular-arithmetic square-numbers
asked 2 days ago
al3ndaleebal3ndaleeb
36
$endgroup$
|
show 3 more comments
$begingroup$
I'm looking for a mathematical solution to find unknown value in the numerator as follows:
IF: $$b_1 + b_2 = 151$$
$$n = 86167$$
$$ (231 * 336 + b_1)^2 mod 86167 = 151^2 $$
$$ (25 * 336 + b_2)^2 mod 86167 = 151^2 $$
So, How can I find $b_1$ and $b_2$ mathematically ??
modular-arithmetic square-numbers
asked 2 days ago
al3ndaleebal3ndaleeb
36
$endgroup$
$begingroup$
What part of this is given, and what part represents your attempt at solving the problem? I also don't know what the term "modular numerator" is supposed to mean.
$endgroup$
– saulspatz
2 days ago
$begingroup$
@EricLee $22801>86167?$
$endgroup$
– saulspatz
2 days ago
1
$begingroup$
Are you allowed to use that $86167=199cdot 433?$ That lets you solve $231cdot 336 b_1equiv pm 151 pmod199$ and $231cdot 336 b_1equivpm 151pmod433$ then solve for four values $b_1$ modulo $86167$ by Chinese remainder theorem. Then similar for $25cdot 336 b_2equiv pm 151pmod199,433$
$endgroup$
– Thomas Andrews
2 days ago
$begingroup$
Is the goal to factor the modulus, or are you know and can use its factorization?
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@ThomasAndrews : Thanks alot for your answer, I was trying to study if this equation will guide me for fast factorization method.
$endgroup$
– al3ndaleeb
2 days ago
|
show 3 more comments
$begingroup$
I'm looking for a mathematical solution to find unknown value in the numerator as follows:
IF: $$b_1 + b_2 = 151$$
$$n = 86167$$
$$ (231 * 336 + b_1)^2 mod 86167 = 151^2 $$
$$ (25 * 336 + b_2)^2 mod 86167 = 151^2 $$
So, How can I find $b_1$ and $b_2$ mathematically ??
modular-arithmetic square-numbers
asked 2 days ago
al3ndaleebal3ndaleeb
36
$endgroup$
I'm looking for a mathematical solution to find unknown value in the numerator as follows:
IF: $$b_1 + b_2 = 151$$
$$n = 86167$$
$$ (231 * 336 + b_1)^2 mod 86167 = 151^2 $$
$$ (25 * 336 + b_2)^2 mod 86167 = 151^2 $$
So, How can I find $b_1$ and $b_2$ mathematically ??
modular-arithmetic square-numbers
modular-arithmetic square-numbers
asked 2 days ago
al3ndaleebal3ndaleeb
36
asked 2 days ago
al3ndaleebal3ndaleeb
36
edited 2 days ago
al3ndaleeb
asked 2 days ago
al3ndaleebal3ndaleeb
36
asked 2 days ago
al3ndaleebal3ndaleeb
36
asked 2 days ago
al3ndaleebal3ndaleeb
36
36
$begingroup$
What part of this is given, and what part represents your attempt at solving the problem? I also don't know what the term "modular numerator" is supposed to mean.
$endgroup$
– saulspatz
2 days ago
$begingroup$
@EricLee $22801>86167?$
$endgroup$
– saulspatz
2 days ago
1
$begingroup$
Are you allowed to use that $86167=199cdot 433?$ That lets you solve $231cdot 336 b_1equiv pm 151 pmod199$ and $231cdot 336 b_1equivpm 151pmod433$ then solve for four values $b_1$ modulo $86167$ by Chinese remainder theorem. Then similar for $25cdot 336 b_2equiv pm 151pmod199,433$
$endgroup$
– Thomas Andrews
2 days ago
$begingroup$
Is the goal to factor the modulus, or are you know and can use its factorization?
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@ThomasAndrews : Thanks alot for your answer, I was trying to study if this equation will guide me for fast factorization method.
$endgroup$
– al3ndaleeb
2 days ago
|
show 3 more comments
$begingroup$
What part of this is given, and what part represents your attempt at solving the problem? I also don't know what the term "modular numerator" is supposed to mean.
$endgroup$
– saulspatz
2 days ago
$begingroup$
@EricLee $22801>86167?$
$endgroup$
– saulspatz
2 days ago
1
$begingroup$
Are you allowed to use that $86167=199cdot 433?$ That lets you solve $231cdot 336 b_1equiv pm 151 pmod199$ and $231cdot 336 b_1equivpm 151pmod433$ then solve for four values $b_1$ modulo $86167$ by Chinese remainder theorem. Then similar for $25cdot 336 b_2equiv pm 151pmod199,433$
$endgroup$
– Thomas Andrews
2 days ago
$begingroup$
Is the goal to factor the modulus, or are you know and can use its factorization?
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@ThomasAndrews : Thanks alot for your answer, I was trying to study if this equation will guide me for fast factorization method.
$endgroup$
– al3ndaleeb
2 days ago
$begingroup$
What part of this is given, and what part represents your attempt at solving the problem? I also don't know what the term "modular numerator" is supposed to mean.
$endgroup$
– saulspatz
2 days ago
$begingroup$
What part of this is given, and what part represents your attempt at solving the problem? I also don't know what the term "modular numerator" is supposed to mean.
$endgroup$
– saulspatz
2 days ago
$begingroup$
@EricLee $22801>86167?$
$endgroup$
– saulspatz
2 days ago
$begingroup$
@EricLee $22801>86167?$
$endgroup$
– saulspatz
2 days ago
1
1
$begingroup$
Are you allowed to use that $86167=199cdot 433?$ That lets you solve $231cdot 336 b_1equiv pm 151 pmod199$ and $231cdot 336 b_1equivpm 151pmod433$ then solve for four values $b_1$ modulo $86167$ by Chinese remainder theorem. Then similar for $25cdot 336 b_2equiv pm 151pmod199,433$
$endgroup$
– Thomas Andrews
2 days ago
$begingroup$
Are you allowed to use that $86167=199cdot 433?$ That lets you solve $231cdot 336 b_1equiv pm 151 pmod199$ and $231cdot 336 b_1equivpm 151pmod433$ then solve for four values $b_1$ modulo $86167$ by Chinese remainder theorem. Then similar for $25cdot 336 b_2equiv pm 151pmod199,433$
$endgroup$
– Thomas Andrews
2 days ago
$begingroup$
Is the goal to factor the modulus, or are you know and can use its factorization?
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
Is the goal to factor the modulus, or are you know and can use its factorization?
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@ThomasAndrews : Thanks alot for your answer, I was trying to study if this equation will guide me for fast factorization method.
$endgroup$
– al3ndaleeb
2 days ago
$begingroup$
@ThomasAndrews : Thanks alot for your answer, I was trying to study if this equation will guide me for fast factorization method.
$endgroup$
– al3ndaleeb
2 days ago
|
show 3 more comments
0
active
oldest
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$begingroup$
What part of this is given, and what part represents your attempt at solving the problem? I also don't know what the term "modular numerator" is supposed to mean.
$endgroup$
– saulspatz
2 days ago
$begingroup$
@EricLee $22801>86167?$
$endgroup$
– saulspatz
2 days ago
1
$begingroup$
Are you allowed to use that $86167=199cdot 433?$ That lets you solve $231cdot 336 b_1equiv pm 151 pmod199$ and $231cdot 336 b_1equivpm 151pmod433$ then solve for four values $b_1$ modulo $86167$ by Chinese remainder theorem. Then similar for $25cdot 336 b_2equiv pm 151pmod199,433$
$endgroup$
– Thomas Andrews
2 days ago
$begingroup$
Is the goal to factor the modulus, or are you know and can use its factorization?
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
@ThomasAndrews : Thanks alot for your answer, I was trying to study if this equation will guide me for fast factorization method.
$endgroup$
– al3ndaleeb
2 days ago