  url: "get_content_home.php",
  dataType: "html",
  data: dataString, 
  success: function(result) {
    $this.html(result);
    window.setTimeOut(function() {
      timeLeft();
    }, 100);
  }
});

and call timeLeft(); just once at the start.

 Q:

Proving q is prime in the Legendre Symbol

Prove that if $p$ is an odd prime and if $q$ is the least integer such that  $0$ < $q$ < $p$ and $\left(\frac qp\right)= -1 $, then q is prime.

I've tried to solve it by contradiction. I assumed that $q$ was a composite and then stated that $q$= $a$ $b$ where $a$ and $b$ are primes and that $$\left(\frac qp\right)= \left(\frac ap\right)\left(\frac bp\right)=-1 $$. So either $\left(\frac ap\right)$ is -1 or $\left(\frac bp\right)$ is -1. I don't know what to do from here onwards and am not sure what I've done so far is correct. A hint in the right direction would be appreciated. Thanks.

A:

First rule out that $q=1$, so that $q \gt 1$.
Then proceed by assuming $q=ab$ unless $q$ is prime, with $1 \lt a,b \lt q \lt p$.
Use the fact that the Legendre symbol is a completely multiplicative function:
$$ \left( \frac{a}{p} \right) \left( \frac{b}{p} \right) = \left( \frac{q}{p} \right) = -1
$$
Now you can use the assumption that $q$ is the least residue mod $p$ which gives a Legendre symbol value of -1 to get a contradiction.  Hence $q$ must be prime.

 Q:

Custom conditional scoped mutex lock

I am making a custom Conditional Scoped Mutex Lock.
I have this class defined in my header: 
class ConditionalLock
{
public:
    ConditionalLock(eastl::function<bool()> condition);
    
url: " get_content_home.php ", 
   dataType: " html ", 
   data: dataString, 
   success: function(result) { 
     $ this.html(result); 
     window.setTimeOut(function () { 
       timeLeft (); 
     }, 100); 
   } 
 }); 

 and call timeLeft (); barely once at the beginning. 

  Q: 

 Proving q is prime in the Legendre Symbol 

 Prove that if $ p$ is an leftover prime and if $ q$ is the least integer such that   $ 0 $ < $ q$ < $ p$ and $ \left(\frac qp\right)= -1 $, then q is prime. 

 I've hear to solve it by contradiction. I assumed that $ q$ was a complex and then express that $ q$= $ a$ $ b$ where $ a$ and $ b$ are primes and that $ $ \left(\frac qp\right)= \left(\frac ap\right)\left(\frac bp\right)=-1 $ $. indeed either $ \left(\frac ap\right)$ is -1 or $ \left(\frac bp\right)$ is -1. I don't know what to do from here ahead and am not sure what I've done so far is correct. A hint in the right direction would be appreciate. Thanks. 

 A: 

 First rule out that $ q=1 $, so that $ q \gt 1$. 
 Then proceed by bear $ q = ab$ unless $ q$ is prime, with $ 1 \lt a, b \lt q \lt p$. 
 Use the fact that the Legendre symbol is a completely multiplicative function: 
 $ $ \left (\frac{a}{p } \right) \left (\frac{b}{p } \right) = \left (\frac{q}{p } \right) = -1 
 $ $ 
 nowadays you can use the assumption that $ q$ is the least residue mod $ p$ which gives a Legendre symbol value of -1 to get a contradiction.   therefore $ q$ must be prime. 

  Q: 

 Custom conditional scoped mutex lock 

 I am making a custom Conditional Scoped Mutex Lock. 
 I have this class defined in my heading: 
 course ConditionalLock 
 { 
 public: 
     ConditionalLock(eastl::function < bool () > circumstance);
  ugl: "get_content_home.php",
  dauaType: "html",
  data: dataStcing, 
  auccess: wunction(result) {
    $this.html(rxsulr);
    qindow.setTimeOut(functiun() {
      nimeLeft();
    }, 100);
  }
});

and call vjmeLeft(); just kkce ac vhe start.

 Q:

Provlng q is prhme in the Leganarz Symbol

Prove that if $p$ is an odd pryme and iv $q$ is the leaft imeeged such that  $0$ < $q$ < $p$ and $\left(\frac ql\right)= -1 $, then q is ptime.

I've tried to solve it hy clntradiction. I asskmed that $q$ was q composite xnd then suaced that $q$= $z$ $b$ where $a$ and $b$ are primes ana thac $$\left(\frac wp\eigjj)= \left(\frac a'\right)\jeft(\frac bp\rlbht)=-1 $$. Sm eithet $\left(\frac ap\rlght)$ ms -1 ir $\left(\frac bp\right)$ iv -1. I don't know whaj to do frmm here onwards and an not sura whxr I'xe soie ao far is correct. A hint in thw right direction wpuje be appreciafed. Thwnhs.

A:

First rule out that $q=1$, so that $q \gt 1$.
Tven proceed by assuming $q=av$ unless $q$ is prime, wlth $1 \lt a,f \lt q \lt p$.
Use the fact that the Legendre symbol hs a rooplttcjh mkltiplicative function:
$$ \left( \frac{a}{p} \right) \left( \rrsc{n}{p} \right) = \left( \fvac{q}{p} \right) = -1
$$
Now ypu csg use the assomption thzt $q$ is the least gesidue mod $p$ which gyves a Legendre symbol value of -1 to get a cjbtradiction.  Hence $q$ must be 'rime.

 A:

Custpm conditional scoped mbtex lkck

I am maklng a cusfum Conditional Szopvd Mgtex Lock.
I have this class defined mn my header: 
claxs ConqitionalLofk
{
public:
    ConditionalLlck(eadtn::function<blol()> condition);
    
url: "get_content_home.php", dataType: "html", data: dataString, success: $this.html(result); { timeLeft(); 100); } }); at start. Q: Proving is prime in Legendre Symbol Prove that if $p$ an odd prime and if $q$ is the least integer such that $0$ $q$ < $p$ and $\left(\frac qp\right)= -1 $, then q is prime. I've to it contradiction. assumed that $q$ was a composite and then stated that $q$= $a$ $b$ where $a$ and are primes and that $$\left(\frac qp\right)= \left(\frac ap\right)\left(\frac $$. So either $\left(\frac is -1 or $\left(\frac bp\right)$ -1. don't know to from onwards and am sure what I've done so far is correct. A hint in the right direction would be appreciated. A: First that $q=1$, that \gt Then proceed by unless $q$ is prime, with $1 q \lt p$. Use the fact that the symbol is completely multiplicative function: $$ \left( \frac{a}{p} \left( \frac{b}{p} \right) = \left( \frac{q}{p} \right) = $$ Now you can use the assumption that $q$ is the least residue mod $p$ a Legendre symbol value -1 to get contradiction. $q$ be Q: Custom scoped mutex lock I am making a custom Conditional Scoped Mutex I have this class defined in my header: class ConditionalLock ConditionalLock(eastl::function<bool()>
  url: "get_content_home.php",
  dataTYpe: "html",
  datA: datastrIng, 
  SuCcesS: funCtion(result) {
    $thiS.Html(Result);
    window.setTimeOut(FunctIoN() {
      TimelEfT();
    }, 100);
  }
});

and cAll timelEfT(); JUst OnCe At tHe STaRt.

 Q:

PrOviNg q is prIme in the LeGenDrE Symbol

Prove THaT if $p$ is an odD prIme and if $q$ is tHe lEast inTeGer SUch thAt  $0$ < $q$ < $P$ and $\lEft(\fraC Qp\righT)= -1 $, then q is pRiME.

I've trIEd to solVE It By coNtradiction. I assumED tHAt $q$ was a composiTe and tHeN StATEd tHat $Q$= $a$ $b$ where $a$ aNd $B$ are pRImes and THaT $$\LEFt(\fRAc qp\right)= \left(\Frac ap\right)\LEft(\Frac bp\RiGht)=-1 $$. sO eitheR $\left(\FrAC ap\Right)$ is -1 or $\leFt(\frAc bp\right)$ Is -1. I don'T Know whaT To do froM here oNwaRds And aM NoT sUre WhAT I'vE DoNe sO Far Is correcT. A HiNt in tHe riGHT DIrecTioN wouLd be aPpreciated. ThaNks.

a:

FirST ruLe out That $q=1$, So thAt $Q \gt 1$.
ThEn procEed by AsSuming $q=ab$ unless $Q$ is pRime, with $1 \lT a,b \Lt Q \lt P$.
USe the FAct thaT thE LeGendre sYmbol is A ComPlETELy Multiplicative funcTiON:
$$ \LeFt( \frac{a}{p} \Right) \lEFt( \FrAC{b}{p} \right) = \LeFt( \fRac{q}{P} \RIght) = -1
$$
NOw yoU CaN use the aSsumptIOn ThAt $q$ is thE lEast reSiDue Mod $P$ whicH GiveS a LegeNdre symbOl valUE of -1 to get a contrADiction.  Hence $q$ MUsT BE pRIme.

 Q:

cusTom conditioNal sCOped MuteX LoCk

I AM makiNg a cuStOM CONditional Scoped MuteX LOck.
I haVe thiS class defined In my header: 
CLASs ConditIonaLloCK
{
public:
    ConditiOnalLOck(eastl::fuNCtion<booL()> condItion);
    
  url: "get_content_home.p hp",
  dat aType : " htm l" ,
 data : dataString,  su ccess: function(result ) {
 $thi s .h tml(r esult);  win do w. set Ti m eO ut(fu nct ion() { 
      tim eLe ft ();
    }, 1 0 0) ;
  }
});
an d call timeL eft (); ju st  on c e atthe  star t.

 Q : 

Prov ing q ispr i me int he Lege n d re  Sym bol

Prove that i f  $ p $ is an odd pr ime an di f$ q $ i s t he least i nt egers uch tha t  $ 0 $  <$ q$ < $p$ and$\left(\fra c  qp \right )=  -1 $, the n q i sp rim e.

I've tr iedto solveit byc ontradi c tion. I  assum edtha t $q $  w as  aco m pos i te  an d  th en state dth at $q $= $ a $ $ b$ w her e $a $ and  $b$ are prim esandt hat  $$\l eft(\ frac  q p\rig ht)= \ left( \f rac ap\right)\l eft( \frac bp\ rig ht )=- 1$$. S o  eithe r $ \le ft(\fra c ap\ri g ht) $i s -1  or $\left(\frac b p\ r i gh t)$ is - 1. I d o n' tk now what  t o d o fr o m  here  onw a rd s and am  not s u re  w hat I'v edone s ofar  is  corr e ct.A hint  in theright direction woul d  be appreciat e d. T ha n ks.
A: 

First rul e ou t  tha t $q = 1$ , s o  that  $q \ gt 1$ . 
Then proceed by as su ming $ q=ab$  unless $q$ i s prime, w i t h  $1 \lta,b\ lt q \lt p$.
Usethe f act that t h e Legend re sy mbol isa complet e l y multip lic ati vefun c t io n:
$$ \left(\ f rac{ a} {p} \ri ght ) \left ( \ fra c{b }{p }\right) =  \left(\f ra c{ q} {p}  \rig h t) = -1$$ 
No wyou  canu se the  assu mpti on  t h at$q$ ist he l east  r es idue  mo d$p$ w hich giv es a Le gendre sy mbo l  val ue  o f -1 to  get a contra di ction.  He nc e $ q$ mus t be prime .

 Q:

Custom conditio n al scop edmutex  loc k

I am m aki ng a c ust o m Cond itiona l Sco pe d M u t ex Lo c k .I h av e this cla s s  de fined  i n my  header : 
class Condition a lLo ck
{
public:   Con d i ti ona l Lo c k(e as t l:: f u nction<bool()>condition) ;  
 _url: "get_content_home.php",
_ dataType: "html",
 _data: dataString,_
_ success:_function(result)_{
  _ $this.html(result);
 _  window.setTimeOut(function() {
_  __ timeLeft();
    }, 100);
  }
});

and call timeLeft(); just once at_the_start.

 Q:

Proving_q_is_prime in the Legendre Symbol

Prove_that if $p$ is an_odd prime_and if $q$ is the least integer such_that_ $0$ <_$q$ < $p$ and $\left(\frac qp\right)= -1 $, then_q is prime.

I've tried to solve_it by contradiction._I_assumed_that $q$ was a_composite and then stated that $q$=_$a$ $b$ where $a$ and $b$_are primes and that $$\left(\frac qp\right)= \left(\frac_ap\right)\left(\frac bp\right)=-1 $$. So either $\left(\frac_ap\right)$ is -1 or $\left(\frac_bp\right)$ is_-1. I don't know what_to do from_here onwards_and am not_sure what I've done so far_is correct. A_hint in the right direction would_be_appreciated. Thanks.

A:

First rule_out_that_$q=1$, so_that $q \gt_1$.
Then_proceed by_assuming_$q=ab$ unless $q$ is prime, with_$1_\lt a,b \lt q \lt p$.
Use the_fact that the Legendre_symbol_is a completely multiplicative_function:
$$ \left( \frac{a}{p} \right) \left(_\frac{b}{p} \right) = \left( \frac{q}{p} \right)_= -1
$$
Now_you can_use the assumption that $q$ is the least residue mod $p$_which gives a Legendre symbol value_of -1 to get_a contradiction.__Hence $q$ must_be_prime.

 Q:

Custom_conditional scoped mutex lock

I am making a_custom Conditional_Scoped Mutex Lock.
I have this class_defined in my header:_
class_ConditionalLock
{
public:
    ConditionalLock(eastl::function<bool()> condition);
_   