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DgrayMan
 Post subject: Conrguence (integers modulo m)
PostPosted: Thu, 26 Apr 2012 17:39:22 UTC 
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If i have two equations ,

11a+b(mod26)=17
&
22a+b(mod26)=22


How do i begin to solve for a and b?
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Shadow
 Post subject: Re: Conrguence (integers modulo m)
PostPosted: Thu, 26 Apr 2012 20:27:29 UTC 
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DgrayMan wrote:
If i have two equations ,

11a+b(mod26)=17
&
22a+b(mod26)=22


How do i begin to solve for a and b?


Use the Chinese remainder theorem.

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DgrayMan
 Post subject: Re: Conrguence (integers modulo m)
PostPosted: Thu, 26 Apr 2012 22:37:38 UTC 
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at first i thought you were messing with me lol
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outermeasure
 Post subject: Re: Conrguence (integers modulo m)
PostPosted: Fri, 27 Apr 2012 05:25:17 UTC 
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Shadow wrote:
DgrayMan wrote:
If i have two equations ,

11a+b(mod26)=17
&
22a+b(mod26)=22


How do i begin to solve for a and b?


Use the Chinese remainder theorem.


I assume the OP means 11a+b\equiv 17\pmod{26} etc. Why apply CRT here instead of eliminating a immediately?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Soroban
 Post subject: Re: Conrguence (integers modulo m)
PostPosted: Fri, 4 May 2012 17:52:27 UTC 
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Hello, DgrayMan!

Quote:
\text{Solve: }\:\begin{array}{cccc} 11a+b & \equiv & 17 \text{ (mod 26)} & [1] \\ 22a+b&\equiv& 22\text{ (mod 26)} & [2] \end{array}

\begin{array}{cccccccc}\text{Multiply [1] by 2:} & 22a + 2b &\equiv & 34\text{ (mod 26)} \\ \text{Multiply [2] by -1:} & \text{-}22a - b &\equiv& -22\text{ (mod 26)} \end{array}

. . \text{Add: }\:b\:\equiv\:12\text{ (mod 26)}


\text{Substitute into [1]: }\:11a + 12 \:\equiv\:17\text{ (mod 26)}

. . . . . . . . . . . . . . . . . . . .11a \:\equiv\:5\text{ (mod 26)}

\text{Multiply by 19: }\qquad\, 19\cdot11a \:\equiv \:19\cdot5\text{ (mod 26)} .**

n . . . . . . . . . . . . . . . . . .209a \:\equiv\:95\text{ (mod 26)}

. . . . . . . . . . . . . . . . . . . . . .a \:\equiv\:17\text{ (mod 26)}


\text{Solution: }\;\begin{Bmatrix}a & \equiv & 17 \text{ (mod 26)} \\ b & \equiv & 12 \text{ (mod 26)} \end{Bmatrix}


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

How do we find that "19"?
Here is a primitive method . . .


We want to find x so that: .11x \:\equiv\:1\text{ (mod 26)}

That is: .11x \:=\:26m + 1

. . Then: .x \:=\:\dfrac{26m+1}{11} \quad\Rightarrow\quad x \:=\:2m + \dfrac{4m+1}{11}\;\;(a)

Since x is an integer, 4m+1 must be a multiple of 11: .4m+1 \:=\:11n

We have: .m \:=\:\dfrac{11n-1}{4} \quad\Rightarrow\quad m \:=\:2n + \dfrac{3n-1}{4}\;\;(b)

Since m is an integer, 3n-1 must be a multiple of 4: .3n-1 \:=\:4p

We have: .n \:=\:\dfrac{4p+1}{3}

By inspection, we find p = 2 \quad\Rightarrow\quad n = 3

Substitute into (b): .m \:=\:2(3) + \dfrac{3(3)-1}{4} \quad\Rightarrow\quad m = 8

Substitute into (a): .x \:=\:2(8) + \dfrac{4(8) + 1}{11} \quad\Rightarrow\quad x \,=\,19


And that is how I found the multiplicative inverse of 11, mod 26.
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outermeasure
 Post subject: Re: Conrguence (integers modulo m)
PostPosted: Sat, 5 May 2012 05:39:56 UTC 
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Soroban wrote:
And that is how I found the multiplicative inverse of 11, mod 26.


Why do it the hard way when you could just use Bezout/Euclid?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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