# reduction mod

• Jun 11th 2008, 02:42 PM
duggaboy
reduction mod
let n=1105, so n-1=2^4(69) Compute the values of

2^69(mod1105), 2^2*69(mod 1105), 2^4*69(mod1105), 2^8*69(mod1105),

Use the Rabin Miller test to conclue that n is composite....

Where do I begin and how?
• Jun 11th 2008, 11:24 PM
CaptainBlack
Quote:

Originally Posted by duggaboy
let $\displaystyle n=1105,$ so $\displaystyle n-1=2^4\times 69$ Compute the values of

$\displaystyle [2^{69} \mod 1105],\ [2^{2 \times 69} \mod 1105],\ [2^{4 \times 69}\mod 1105],\ [2^{8 \times 69} \mod 1105]$

Use the Rabin Miller test to conclue that n is composite....

Where do I begin and how?

In the Rabin-Miller test for the (pseudo) primality of $\displaystyle 1105$ we have $\displaystyle s=4$ and $\displaystyle d=69$

So $\displaystyle a$ is a witness to the primallity of $\displaystyle 1105$ if:

$\displaystyle a^d \not \equiv 1 \mod 1105$

and

$\displaystyle a^{2^r.d} \not \equiv -1 \mod 1105; \forall \ r \in \{0,\ ..., \ s-1\}$

So if you compute the values asked for and the first is not congruent to $\displaystyle \pm 1 \mod 1105$ and the others are not congruent to $\displaystyle -1 \mod 1105$, then $\displaystyle 2$ is a witness to the compositness of $\displaystyle 1105$. (In fact if the conditions hold then 1105 is certainly composite, if they fail 1105 is probably prime)

RonL
• Jun 12th 2008, 10:53 AM
duggaboy
http://www.mathhelpforum.com/math-he...c4468b91-1.gif
Can you explain the last part of this to me?
• Jun 12th 2008, 02:05 PM
CaptainBlack
Quote:

Originally Posted by duggaboy
http://www.mathhelpforum.com/math-he...c4468b91-1.gif
Can you explain the last part of this to me?

It is that these:

"2^69(mod1105), 2^2*69(mod 1105), 2^4*69(mod1105), 2^8*69(mod1105)"

are not congruent to -1 modulo 1105.

RonL