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Prime number raised to a power


question on prime numberPrime number questionFind the prime-power decomposition of 999999999999If all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Is the square of a prime necessarily larger than the next prime?There are infinitely many odd numbers not expressible as the sum of a prime number and a power of $2$Is there any deep reason that 23456789 is prime?Is this number prime?Advise on determining prime numbers such that the product of their geometric series is equivalent to another prime number raised to a power.Is there a name for a multiplicative function $psi(p^n) = (p^n-1)(p^n-1-1)cdots(p-1)$?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago

















2












$begingroup$


Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago













2












2








2


1



$begingroup$


Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$




Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 8 hours ago









Math GuyMath Guy

1807 bronze badges




1807 bronze badges







  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago












  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago







2




2




$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago




$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago










5 Answers
5






active

oldest

votes


















4












$begingroup$

We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you! Which part uses specifically that $p$ is a prime number?
    $endgroup$
    – Math Guy
    8 hours ago






  • 2




    $begingroup$
    This does not depend on $p$ prime. It works for any $n$.
    $endgroup$
    – Antonios-Alexandros Robotis
    8 hours ago






  • 2




    $begingroup$
    Nowhere, we do not need prime numbers here.
    $endgroup$
    – Azif00
    8 hours ago


















4












$begingroup$

We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



Otherwise, let me prove it for you here (at least for you to compare):



Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    If $n = 1$, we have $p mid a$ and we are done.



    For $n ge 2$:



    $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



    This result, being so simple, is very well-known.



    We note that $p$ needn't be prime.






    share|cite|improve this answer











    $endgroup$




















      2












      $begingroup$

      Maybe this is an easier way to look at this:
      For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






      share|cite|improve this answer









      $endgroup$




















        0












        $begingroup$

        If $p^n | a$ then $a = p^nk$ for some integer $k$.



        As $p | p^kk$, so we have $p|a$.






        share|cite|improve this answer









        $endgroup$















          Your Answer








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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago















          4












          $begingroup$

          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago













          4












          4








          4





          $begingroup$

          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






          share|cite|improve this answer









          $endgroup$



          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Azif00Azif00

          9479 bronze badges




          9479 bronze badges











          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago
















          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago















          $begingroup$
          Thank you! Which part uses specifically that $p$ is a prime number?
          $endgroup$
          – Math Guy
          8 hours ago




          $begingroup$
          Thank you! Which part uses specifically that $p$ is a prime number?
          $endgroup$
          – Math Guy
          8 hours ago




          2




          2




          $begingroup$
          This does not depend on $p$ prime. It works for any $n$.
          $endgroup$
          – Antonios-Alexandros Robotis
          8 hours ago




          $begingroup$
          This does not depend on $p$ prime. It works for any $n$.
          $endgroup$
          – Antonios-Alexandros Robotis
          8 hours ago




          2




          2




          $begingroup$
          Nowhere, we do not need prime numbers here.
          $endgroup$
          – Azif00
          8 hours ago




          $begingroup$
          Nowhere, we do not need prime numbers here.
          $endgroup$
          – Azif00
          8 hours ago













          4












          $begingroup$

          We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



          Otherwise, let me prove it for you here (at least for you to compare):



          Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






          share|cite|improve this answer









          $endgroup$

















            4












            $begingroup$

            We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



            Otherwise, let me prove it for you here (at least for you to compare):



            Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






            share|cite|improve this answer









            $endgroup$















              4












              4








              4





              $begingroup$

              We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



              Otherwise, let me prove it for you here (at least for you to compare):



              Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






              share|cite|improve this answer









              $endgroup$



              We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



              Otherwise, let me prove it for you here (at least for you to compare):



              Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 8 hours ago









              ThorWittichThorWittich

              1,8061 silver badge12 bronze badges




              1,8061 silver badge12 bronze badges





















                  3












                  $begingroup$

                  If $n = 1$, we have $p mid a$ and we are done.



                  For $n ge 2$:



                  $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                  This result, being so simple, is very well-known.



                  We note that $p$ needn't be prime.






                  share|cite|improve this answer











                  $endgroup$

















                    3












                    $begingroup$

                    If $n = 1$, we have $p mid a$ and we are done.



                    For $n ge 2$:



                    $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                    This result, being so simple, is very well-known.



                    We note that $p$ needn't be prime.






                    share|cite|improve this answer











                    $endgroup$















                      3












                      3








                      3





                      $begingroup$

                      If $n = 1$, we have $p mid a$ and we are done.



                      For $n ge 2$:



                      $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                      This result, being so simple, is very well-known.



                      We note that $p$ needn't be prime.






                      share|cite|improve this answer











                      $endgroup$



                      If $n = 1$, we have $p mid a$ and we are done.



                      For $n ge 2$:



                      $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                      This result, being so simple, is very well-known.



                      We note that $p$ needn't be prime.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 8 hours ago

























                      answered 8 hours ago









                      Robert LewisRobert Lewis

                      50.8k2 gold badges33 silver badges69 bronze badges




                      50.8k2 gold badges33 silver badges69 bronze badges





















                          2












                          $begingroup$

                          Maybe this is an easier way to look at this:
                          For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






                          share|cite|improve this answer









                          $endgroup$

















                            2












                            $begingroup$

                            Maybe this is an easier way to look at this:
                            For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






                            share|cite|improve this answer









                            $endgroup$















                              2












                              2








                              2





                              $begingroup$

                              Maybe this is an easier way to look at this:
                              For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






                              share|cite|improve this answer









                              $endgroup$



                              Maybe this is an easier way to look at this:
                              For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 8 hours ago









                              Ruben du BurckRuben du Burck

                              9432 silver badges14 bronze badges




                              9432 silver badges14 bronze badges





















                                  0












                                  $begingroup$

                                  If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                  As $p | p^kk$, so we have $p|a$.






                                  share|cite|improve this answer









                                  $endgroup$

















                                    0












                                    $begingroup$

                                    If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                    As $p | p^kk$, so we have $p|a$.






                                    share|cite|improve this answer









                                    $endgroup$















                                      0












                                      0








                                      0





                                      $begingroup$

                                      If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                      As $p | p^kk$, so we have $p|a$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                      As $p | p^kk$, so we have $p|a$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 7 hours ago









                                      mlchristiansmlchristians

                                      44414 bronze badges




                                      44414 bronze badges



























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Data for the year 2010.„Israel's accession to the OECD”Popular Opinion„On the Move”Hosea 12:5„Walking the Bible Timeline”„Palestine: History”„Return to Zion”An invention called 'the Jewish people' – Haaretz – Israel NewsoriginalJewish and Non-Jewish Population of Palestine-Israel (1517–2004)ImmigrationJewishvirtuallibrary.orgChapter One: The Heralders of Zionism„The birth of modern Israel: A scrap of paper that changed history”„League of Nations: The Mandate for Palestine, 24 iulie 1922”The Population of Palestine Prior to 1948originalBackground Paper No. 47 (ST/DPI/SER.A/47)History: Foreign DominationTwo Hundred and Seventh Plenary Meeting„Israel (Labor Zionism)”Population, by Religion and Population GroupThe Suez CrisisAdolf EichmannJustice Ministry Reply to Amnesty International Report„The Interregnum”Israel Ministry of Foreign Affairs – The Palestinian National Covenant- July 1968Research on terrorism: trends, achievements & failuresThe Routledge Atlas of the Arab–Israeli conflict: The Complete History of the Struggle and the Efforts to Resolve It"George Habash, Palestinian Terrorism Tactician, Dies at 82."„1973: Arab states attack Israeli forces”Agranat Commission„Has Israel Annexed East Jerusalem?”original„After 4 Years, Intifada Still Smolders”From the End of the Cold War to 2001originalThe Oslo Accords, 1993Israel-PLO Recognition – Exchange of Letters between PM Rabin and Chairman Arafat – Sept 9- 1993Foundation for Middle East PeaceSources of Population Growth: Total Israeli Population and Settler Population, 1991–2003original„Israel marks Rabin assassination”The Wye River Memorandumoriginal„West Bank barrier route disputed, Israeli missile kills 2”"Permanent Ceasefire to Be Based on Creation Of Buffer Zone Free of Armed Personnel Other than UN, Lebanese Forces"„Hezbollah kills 8 soldiers, kidnaps two in offensive on northern border”„Olmert confirms peace talks with Syria”„Battleground Gaza: Israeli ground forces invade the strip”„IDF begins Gaza troop withdrawal, hours after ending 3-week offensive”„THE LAND: Geography and Climate”„Area of districts, sub-districts, natural regions and lakes”„Israel - 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Ethnic groups”„Jews, by country of origin and age”„Minority Communities in Israel: Background & Overview”„Israel”„Language in Israel”„Selected Data from the 2011 Social Survey on Mastery of the Hebrew Language and Usage of Languages”„Religions”„5 facts about Israeli Druze, a unique religious and ethnic group”„Israël”Israel Country Study Guide„Haredi city in Negev – blessing or curse?”„New town Harish harbors hopes of being more than another Pleasantville”„List of localities, in alphabetical order”„Muncitorii români, doriți în Israel”„Prietenia româno-israeliană la nevoie se cunoaște”„The Higher Education System in Israel”„Middle East”„Academic Ranking of World Universities 2016”„Israel”„Israel”„Jewish Nobel Prize Winners”„All Nobel Prizes in Literature”„All Nobel Peace Prizes”„All Prizes in Economic Sciences”„All Nobel Prizes in Chemistry”„List of Fields Medallists”„Sakharov Prize”„Țara care și-a sfidat "destinul" și se bate umăr la umăr cu Silicon Valley”„Apple's R&D center in Israel grew to about 800 employees”„Tim Cook: Apple's Herzliya R&D center second-largest in world”„Lecții de economie de la Israel”„Land use”Israel Investment and Business GuideA Country Study: IsraelCentral Bureau of StatisticsFlorin Diaconu, „Kadima: Flexibilitate și pragmatism, dar nici un compromis în chestiuni vitale", în Revista Institutului Diplomatic Român, anul I, numărul I, semestrul I, 2006, pp. 71-72Florin Diaconu, „Likud: Dreapta israeliană constant opusă retrocedării teritoriilor cureite prin luptă în 1967", în Revista Institutului Diplomatic Român, anul I, numărul I, semestrul I, 2006, pp. 73-74MassadaIsraelul a crescut in 50 de ani cât alte state intr-un mileniuIsrael Government PortalIsraelIsraelIsraelmmmmmXX451232cb118646298(data)4027808-634110000 0004 0372 0767n7900328503691455-bb46-37e3-91d2-cb064a35ffcc1003570400564274ge1294033523775214929302638955X146498911146498911

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                                          Mentenanță CS1: Dată și anMentenanță CS1: Utilizare explicită a lui et al.IsraelȚări în Orientul MijlociuRepubliciSionismState membre ale Organizației Națiunilor Unite ebraicăarabărepublică parlamentarăOrientul MijlociuMării MediteraneLibanulSiriaIordaniaCisiordaniaEgiptFâșia GazaIerusalimTel AvivOrganizației Națiunilor Unite14 mai1948Mandatului britanic pentru PalestinaDavid Ben-GurionOrganizației SionisteAgenției Evreiești pentru PalestinaStatului IsraelONULiga ArabăpalestinieneRăzboiul arabo-israelian din 1948Conflict arabo-israelianTransiordaniaEgiptulRăzboiului de Șase ZileCisiordaniaPeninsula SinaiFâșia GazaÎnălțimile GolanBiroului Central de Statisticădruzicerchezisamariteniarmenidemocrație reprezentativăsistem parlamentarvot universalKnessetulgaze naturaleIndicelui Dezvoltării UmaneOCDEȚara I...
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                                          Кастелфранко ди Сопра Становништво Референце Спољашње везе Мени за навигацију43°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.5588543°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.558853179688„The GeoNames geographical database”„Istituto Nazionale di Statistica”проширитиууWorldCat156923403n850174324558639-1cb14643287r(подаци)

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