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\newtheorem*{solution}{\c C\"oz\"um}
\newtheorem*{remark}{Not}

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\begin{document}
%\subtitle{S\i nav I}
\title{Say\i lar Kuram\i na Giri\c s}
\subtitle{{\scshape MAT} \newstylenums{111}, Final S\i nav\i\
\itshape \c c\"oz\"umleri}
\author{David Pierce}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U}
\date{2 Ocak 2018}
%\publishers{Matematik B\"ol\"um\"u, MSGS\"U}
%\mbox{}\\
%\url{dpierce@msgsu.edu.tr}\\
%\url{http://mat.msgsu.edu.tr/~dpierce/}

\maketitle\thispagestyle{empty}

\begin{itshape}
\c C\"oz\"um y\"ontemlerinizi d\"u\c s\"unerek se\c cin
ve \c c\"oz\"umlerinizi net bir \c sekilde yaz\i n.
M\"umk\"unse cevaplar\i n\i z\i\ kontrol edin.
Yazd\i\u g\i n\i z her c\"umlenin, okura a\c c\i k olan nedeni olmal\i.
	
Her s\i navdaki gibi,
ba\c ska kimseyle konu\c smay\i n;
ba\c ska kimsenin ka\u g\i d\i n\i\ g\"ormeye \c cal\i\c smay\i n;
kendi ka\u g\i d\i n\i z\i\ ba\c ska kimseye g\"ostermeyin;
getirdi\u giniz notlar\i, kitaplar\i\ veya cihazlar\i\ kullanmay\i n.

\.Iyi \c cal\i\c smalar dilerim.
\end{itshape}
\begin{problem}
  \"Ozyineli tan\i ma g\"ore
  \begin{align*}
    F_1&=1,&F_2&=1,&F_{n+2}&=F_{n+1}+F_n,\\
    L_1&=1,&L_2&=3,&L_{n+2}&=L_{n+1}+L_n
  \end{align*}
olsun.  Her $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
  L_{n+2}\cdot F_n-L_{n+1}\cdot F_{n+1}=(-1)^{n+1}
\end{equation*}
e\c sitli\u gini g\"osterin.
\end{problem}

\begin{solution}
  \begin{asparaenum}[(i)]
    \item
  $L_3\cdot F_1-L_2\cdot F_2=4-3=1=(-1)^2$ oldu\u gundan
iddia $n=1$ durumunda do\u grudur.
\item
Bir $m$ i\c cin iddia $n=m$ durumunda do\u gru ise
\begin{align*}
&\phantom{{}={}}L_{m+3}\cdot F_{m+1}-L_{m+2}\cdot F_{m+2}\\
&=(L_{m+2}+L_{m+1})\cdot F_{m+1}-L_{m+2}\cdot(F_{m+1}+F_m)\\
&=L_{m+1}\cdot F_{m+1}-L_{m+2}\cdot F_m\\
&=-(-1)^{m+1}\\
&=(-1)^{m+2},  
\end{align*}
dolay\i s\i yla iddia $n=m+1$ durumunda do\u grudur.
\end{asparaenum}
T\"umevar\i mdan her $n$ sayma say\i s\i\ i\c cin iddia do\u grudur.
\end{solution}



\begin{problem}
  A\c sa\u g\i daki kalanda\c sl\i k sisteminin 
en k\"u\c c\"uk pozitif \c c\"oz\"um\"un\"u hesaplay\i n.
\begin{align*}
  x&\equiv 20\pmod{47},&
  x&\equiv 24\pmod{115}.
\end{align*}
\emph{Not}: $47\cdot 115=5405$.
\end{problem}

\begin{solution}
  \"Oklid Algoritmas\i\ ile
  \begin{align*}
    115&=47\cdot2+21,\\
47&=21\cdot2+5\\
21&=5\cdot4+1,
  \end{align*}
dolay\i s\i yla
\begin{align*}
  1
&=21-5\cdot 4=21-(47-21\cdot2)\cdot4\\
&=21\cdot9-47\cdot4=(115-47\cdot2)\cdot9-47\cdot4\\
&=115\cdot9-47\cdot22.
\end{align*}
Kontrol ederiz:
\begin{gather*}
  115\cdot9=1150-115=1035,\\
47\cdot22=940+94=1034.
\end{gather*}
\c Cin Kalan Teoremi'nden sistemin \c c\"oz\"um\"u,
$5405$'e g\"ore
\begin{align*}
  x
&\equiv20\cdot115\cdot9-24\cdot47\cdot22\\
&\equiv20-4\cdot47\cdot22\\
&\equiv20-4\cdot1034\\
&\equiv20-4136\\
&\equiv-4116\\
&\equiv5405-4116\\
&\equiv1289.
\end{align*}
Kontrol ederiz:
\begin{align*}
&  \begin{gathered}
    47\cdot20=940,\\
1289-940=349,\\
47\cdot7=350-21=329,\\
349-329=20,\\
1289\equiv20\pmod{47},
  \end{gathered}&
\begin{gathered}
115\cdot11=1150+115=1265,\\
1289-1265=24,\\
1289\equiv24\pmod{115}.             
\end{gathered}
\end{align*}
\end{solution}



\begin{problem}
  $p=2^{2^4}+1=2^{16}+1=65537$ olsun;
$p$'nin asal oldu\u gunu biliyoruz.
%(Asl\i nda $2^{2^n}+1$ bi\c cimli asallar,
%\textbf{Fermat asallar\i d\i r,}
%ama bildi\u gimiz kadar\i\ ile $p$, bunlar\i n en b\"uy\"u\u g\"ud\"ur.)
Ayr\i ca
\begin{equation*}
 3^{2^{15}}\equiv-1\pmod p.
\end{equation*}
%(Bunu bir ``spreadsheet'' ile kontrol ettim.)
A\c sa\u g\i daki \"onermeleri kan\i tlay\i n.
\begin{enumerate}[(a)]
\item 
$2$, $p$'nin ilkel bir k\"ok\"u de\u gildir.
\item
$3$, $p$'nin ilkel bir k\"ok\"ud\"ur.
\end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[(a)]
    \item
$p$'ye g\"ore
$2^{2^4}\equiv-1$ oldu\u gundan
\begin{equation*}
  2^{2^5}\equiv(2^{2^4})^2\equiv(-1)^2\equiv1,
\end{equation*}
ama $2^5<2^{16}=p-1$, dolay\i s\i yla
$2$'nin mertebesi, $p-1$'den kesin k\"u\c c\"ukt\"ur.
\item
$p$'ye g\"ore $3$'\"un mertebesi $d$ olsun.
O zaman $d\divides2^{16}$,
dolay\i s\i yla bir $t$ i\c cin $d=2^t$ ve $0\leqslant t\leqslant 16$.
E\u ger $t<16$ ise, o zaman $t\leqslant15$, dolay\i s\i yla
\begin{equation*}
-1\equiv3^{2^{15}}\equiv(3^{2^t})^{2^{15-t}}\equiv1^{2^{15-t}}\equiv1,
\end{equation*}
ki bu imk\^ans\i zd\i r.
Sonu\c c olarak $t=16$ ve $d=p-1$,
dolay\i s\i yla $3$, $p$'nin ilkel bir k\"ok\"ud\"ur.
\end{enumerate}
\end{solution}



\begin{problem}
  A\c sa\u g\i daki gerektirmelerin her birini ya kan\i tlay\i n
ya da kar\c s\i t \"ornek ile \c c\"ur\"ut\"un.
(K\"u\c c\"uk harfler tamsay\i d\i r.)
\begin{enumerate}[(a)]
\item
$b\ndivides a$ ise $a\divides b$.
\item 
$\gcd(a,b)=\gcd(a,c)$ ise $\gcd(a,b)=\gcd(a,b+c)$.
\item 
$\gcd(a,b)=1\And\gcd(a,c)=1$ ise $\gcd(a,bc)=1$.
\end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}[(a)]
    \item
$2\ndivides 3\And 3\ndivides 2$.
\item
$\gcd(2,1)=1$ ama $\gcd(2,1+1)=2$.
\item
$\gcd(a,bc)\neq1$ olsun.
O zaman $a$ ve $bc$'nin $p$ \emph{asal} ortak b\"oleni vard\i r.
\"Oklid Lemmas\i\ ile $p\divides b$ veya $p\divides c$.
Bu durumda $p\divides a$ da oldu\u gundan $\gcd(a,b)$
ve $\gcd(a,c)$'den biri $1$ olmaz.
\end{enumerate}
\end{solution}

\end{document}
