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MATHmaniac aka CrUNchy

"Mathematics is the Queen of Science & Number Theory is the Queen of Mathematics"

XM mud....argggghhh

| at 10:27 PM

For 082,,,,gud lux weh,
make sure we scor *NUMBER THOERY*

STUDY=X FAIL
X STUDY=FAIL
(STUDY+X STUDY)=(FAIL+X FAIL)
STUDY(1+X)=FAIL(1+X)
HENCE,
STUDY=FAIL

WARNING!!!!
(DUN TRY THIS DURING XM)

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THANK YOU,SIR

To all our lecturers,Dr Jesni,Prof Torla and of course,Br Supian..
Thank you so much for teaching us this whole sem,.
We also want to apologize for any mistakes that we have done,
You guys are not only our lecturers but also a brother and father to us..
TERIMA KASIH,CIKGU!!
CTS ROCK!!

Same to all friends,
Sorry,Thank You and Good Luck in everything!!

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Pell's Equation

Labels: Exercises | at 8:28 PM

Solve $x^{2}-18y^{2}= 1$ .Give 3 positive solutions.

First step:
Find the continued expansion of $\sqrt{18}$

$x_{0}:\sqrt{18}=4+(\sqrt{18}-4), a_{0}=4$
$x_{1}:\frac{1}{\sqrt{18}-4}.\frac{\sqrt{18}+4}{\sqrt{18}+4}=\frac{8+(\sqrt{18}-4)}{2}=4+\frac{\sqrt{18}-4}{2},a_{1}=4$
$x_{2}:\frac{2}{\sqrt{18}-4}.\frac{\sqrt{18}+4}{\sqrt{18}+4}=\frac{2(\sqrt{18}+4)}{2}=8+\sqrt{18}-4,a_{2}=8$
thus, $\sqrt{18}=\left [ 4;4,8 \right ]$

Second step:
Since n=2 and it is even, the solutions are
$x=P_{2k-1}, y=q_{2k-1}$ where k=1,2,3...
Thus,
$x= P_{1}, y=P_{1}$
$x= P_{3}, y=P_{3}$
$x=p_{5},y=q_{5}$
.
.
.
are the positive solutions.

Third step:
Calculate the value of the numerator and denominator,
$C_{0}=\frac{P_{0}}{q_{0}}=4$
$C_{1}=\frac{P_{1}}{q_{1}}=a_{0}+\frac{1}{a_{1}}=4+\frac{1}{4}=\frac{17}{4}$
$C_{2}=\frac{P_{2}}{q_{2}}=\frac{a_{2}p_{1}+p_{2}}{a_{2}q_{1}+q_{2}}=\frac{8(17)+4}{8(4)+1}=\frac{140}{33}$
$C_{3}=\frac{P_{3}}{q_{3}}=\frac{a_{3}p_{2}+p_{1}}{a_{3}q_{2}+q_{1}}=\frac{4(140)+17}{4(33)+4}=\frac{577}{136}$
$C_{4}=\frac{P_{4}}{q_{4}}=\frac{a_{4}p_{3}+p_{2}}{a_{4}q_{3}+q_{2}}=\frac{8(577)+140}{8(136)+33}=\frac{4756}{1121}$
$C_{5}=\frac{P_{5}}{q_{5}}=\frac{a_{5}p_{4}+p_{3}}{a_{5}q_{4}+q_{3}}=\frac{4(4756)+577}{4(1121)+136}=\frac{19601}{4620}$

So,the answers for 3 positive solutions of $x^{2}-18y^{2}= 1$ are

$\frac{17}{4}$ , $\frac{577}{136}$ , $\frac{19601}{4620}$

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Fermat's Theorem

Labels: Example, Theorem | at 7:28 PM

THEOREM 1
Let P be a prime number and suppose that P is not divisible by a.Then, $a^{p-1}\equiv 1(mod P)$

THEOREM 2
If P is a prime number,then $a^{p}\equiv a(mod P)$ for any integer a.

Now,Using Fermat's Theorem we're going to proof:

a) $1^{P-1}+2^{P-1}+3^{P-1}+...+(P-1)\equiv -1(mod P)$
Theorem 1 said, $a^{p-1}\equiv 1(mod P)$ so,
$1^{P-1}\equiv 1(modP)$
$2^{P-1}\equiv 1(modP)$
$3^{P-1}\equiv 1(modP)$
.
.
.
$(P-1)^{P-1}\equiv 1(modP)$
thus,
$1^{P-1}+2^{P-1}+3^{P-1}+...+(P-1)^{P-1}\equiv (P-1)(modP)$
simplify it,
$1^{P-1}+2^{P-1}+3^{P-1}+...+(P-1)^{P-1}\equiv -1(modP)$
How (P-1) become congruence to -1??
Ok,let's say P as 19
so,when (19-1) become 18,then
$18\equiv -1(mod 19)$

b) $1^{P}+2^{P}+3^{P}+...+(P-1)^{P}\equiv 0(modP)$
THEOREM 2 said,
$a^{p}\equiv a(mod P)$ so,
$1^{P} \equiv 1(modP)$
$2^{P} \equiv 2(modP)$
$3^{P} \equiv 3(modP)$
.
.
.
$(P-1)^{P} \equiv (P-1)(modP)$
thus,
$1^{P}+2^{P}+3^{P}+...+(P-1)^{P}\equiv ((P(P-1))\div (2) )(mod P)$
simplify it,
$1^{P}+2^{P}+3^{P}+...+(P-1)^{P}\equiv 0(mod P)$
How come $((P(P-1)\div (2))\equiv 0(mod P)$
Again P as 19,thus $((19(18)\div (2))\equiv 0(mod P)$

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