For $a$, $b$, $c$ distinct integers, and $P$ a polynomial with integer coefficients, $P(a)=b$, $P(b)=c$, $P(c)=a$ cannot be satisfied simultaneously

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I have pasted the problem and its solution below. My question is about (1), (2), and (3). What does $P_1(x)$, $P_2(x)$, and $P_3(x)$ represent? Near the end they say it's a positive integer. So couldn't we just put $k$ there instead, where $k$ represents some positive integer?



Thanks



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contest-math proof-explanation




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    $P(x)$ is the polynomial that purportedly satisfies the conditions. $P_1(x)$, $P_2(x)$, and $P_3(x)$ are the polynomials left over after factoring out a linear term (we know at least one zero by hypothesis).
    – Clayton
    6 hours ago











  • It says that the quantity $P_1(c)$ obtained by evaluating the polynomial $P_1(x)$ at the integer $c$ is an integer. $P_1(x), P_2(x)$ and $P_3(x)$ are polynomials in $x$ obtained from the original polynomial $P(x)$ using the given data and you can't just replace any of them by the integer $k$.
    – Rob Arthan
    5 hours ago











  • @clayton. Thanks, that clears it up.
    – john fowles
    5 hours ago














up vote
0
down vote

favorite












I have pasted the problem and its solution below. My question is about (1), (2), and (3). What does $P_1(x)$, $P_2(x)$, and $P_3(x)$ represent? Near the end they say it's a positive integer. So couldn't we just put $k$ there instead, where $k$ represents some positive integer?



Thanks



enter image description here



enter image description here




contest-math proof-explanation




share|cite|improve this question

















  • 2




    $P(x)$ is the polynomial that purportedly satisfies the conditions. $P_1(x)$, $P_2(x)$, and $P_3(x)$ are the polynomials left over after factoring out a linear term (we know at least one zero by hypothesis).
    – Clayton
    6 hours ago











  • It says that the quantity $P_1(c)$ obtained by evaluating the polynomial $P_1(x)$ at the integer $c$ is an integer. $P_1(x), P_2(x)$ and $P_3(x)$ are polynomials in $x$ obtained from the original polynomial $P(x)$ using the given data and you can't just replace any of them by the integer $k$.
    – Rob Arthan
    5 hours ago











  • @clayton. Thanks, that clears it up.
    – john fowles
    5 hours ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have pasted the problem and its solution below. My question is about (1), (2), and (3). What does $P_1(x)$, $P_2(x)$, and $P_3(x)$ represent? Near the end they say it's a positive integer. So couldn't we just put $k$ there instead, where $k$ represents some positive integer?



Thanks



enter image description here



enter image description here




contest-math proof-explanation




share|cite|improve this question













I have pasted the problem and its solution below. My question is about (1), (2), and (3). What does $P_1(x)$, $P_2(x)$, and $P_3(x)$ represent? Near the end they say it's a positive integer. So couldn't we just put $k$ there instead, where $k$ represents some positive integer?



Thanks



enter image description here



enter image description here





contest-math proof-explanation





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Batominovski

22.2k22675




22.2k22675









asked 6 hours ago









john fowles

1,047716




1,047716







  • 2




    $P(x)$ is the polynomial that purportedly satisfies the conditions. $P_1(x)$, $P_2(x)$, and $P_3(x)$ are the polynomials left over after factoring out a linear term (we know at least one zero by hypothesis).
    – Clayton
    6 hours ago











  • It says that the quantity $P_1(c)$ obtained by evaluating the polynomial $P_1(x)$ at the integer $c$ is an integer. $P_1(x), P_2(x)$ and $P_3(x)$ are polynomials in $x$ obtained from the original polynomial $P(x)$ using the given data and you can't just replace any of them by the integer $k$.
    – Rob Arthan
    5 hours ago











  • @clayton. Thanks, that clears it up.
    – john fowles
    5 hours ago












  • 2




    $P(x)$ is the polynomial that purportedly satisfies the conditions. $P_1(x)$, $P_2(x)$, and $P_3(x)$ are the polynomials left over after factoring out a linear term (we know at least one zero by hypothesis).
    – Clayton
    6 hours ago











  • It says that the quantity $P_1(c)$ obtained by evaluating the polynomial $P_1(x)$ at the integer $c$ is an integer. $P_1(x), P_2(x)$ and $P_3(x)$ are polynomials in $x$ obtained from the original polynomial $P(x)$ using the given data and you can't just replace any of them by the integer $k$.
    – Rob Arthan
    5 hours ago











  • @clayton. Thanks, that clears it up.
    – john fowles
    5 hours ago







2




2




$P(x)$ is the polynomial that purportedly satisfies the conditions. $P_1(x)$, $P_2(x)$, and $P_3(x)$ are the polynomials left over after factoring out a linear term (we know at least one zero by hypothesis).
– Clayton
6 hours ago





$P(x)$ is the polynomial that purportedly satisfies the conditions. $P_1(x)$, $P_2(x)$, and $P_3(x)$ are the polynomials left over after factoring out a linear term (we know at least one zero by hypothesis).
– Clayton
6 hours ago













It says that the quantity $P_1(c)$ obtained by evaluating the polynomial $P_1(x)$ at the integer $c$ is an integer. $P_1(x), P_2(x)$ and $P_3(x)$ are polynomials in $x$ obtained from the original polynomial $P(x)$ using the given data and you can't just replace any of them by the integer $k$.
– Rob Arthan
5 hours ago





It says that the quantity $P_1(c)$ obtained by evaluating the polynomial $P_1(x)$ at the integer $c$ is an integer. $P_1(x), P_2(x)$ and $P_3(x)$ are polynomials in $x$ obtained from the original polynomial $P(x)$ using the given data and you can't just replace any of them by the integer $k$.
– Rob Arthan
5 hours ago













@clayton. Thanks, that clears it up.
– john fowles
5 hours ago




@clayton. Thanks, that clears it up.
– john fowles
5 hours ago










2 Answers
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up vote
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accepted










Here $P_1$, $P_2$, and $P_3$ each represent a polynomial (with integer coefficients), and equations (1,2,3) are equations of polynomials in $x$ (not just single numbers). For instance, $P_1(x)$ is the polynomial you obtain by doing polynomial long division to divide $P(x)-b$ by $x-a$ (there will be no remainder since $P(a)=b$).



In particular, if you plug in any integer value for $x$, these polynomials will give integer outputs. That's why $P_1(c)$ is an integer.






share|cite|improve this answer




























    up vote
    1
    down vote













    From the given condition the polynomial $P(x)$ can be factorized as follow



    • $P(a)-b=0 implies P(x)-b=P_1(x)cdot (x-a)$

    where $P_1(x)$ is a polynomial of degree $n-1$ (assuming $P(x)$ is of degree $n$).



    Since $P_1(c)neq 0$ is an integer we have $|P_1(c)|ge 1$.



    Note that $P_1(c)neq 0$ since



    • $P(x)-b=P_1(x)cdot (x-a)implies P(c)-b=a-b=P_1(c)cdot (c-a)neq 0$





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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      active

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      up vote
      1
      down vote



      accepted










      Here $P_1$, $P_2$, and $P_3$ each represent a polynomial (with integer coefficients), and equations (1,2,3) are equations of polynomials in $x$ (not just single numbers). For instance, $P_1(x)$ is the polynomial you obtain by doing polynomial long division to divide $P(x)-b$ by $x-a$ (there will be no remainder since $P(a)=b$).



      In particular, if you plug in any integer value for $x$, these polynomials will give integer outputs. That's why $P_1(c)$ is an integer.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Here $P_1$, $P_2$, and $P_3$ each represent a polynomial (with integer coefficients), and equations (1,2,3) are equations of polynomials in $x$ (not just single numbers). For instance, $P_1(x)$ is the polynomial you obtain by doing polynomial long division to divide $P(x)-b$ by $x-a$ (there will be no remainder since $P(a)=b$).



        In particular, if you plug in any integer value for $x$, these polynomials will give integer outputs. That's why $P_1(c)$ is an integer.






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Here $P_1$, $P_2$, and $P_3$ each represent a polynomial (with integer coefficients), and equations (1,2,3) are equations of polynomials in $x$ (not just single numbers). For instance, $P_1(x)$ is the polynomial you obtain by doing polynomial long division to divide $P(x)-b$ by $x-a$ (there will be no remainder since $P(a)=b$).



          In particular, if you plug in any integer value for $x$, these polynomials will give integer outputs. That's why $P_1(c)$ is an integer.






          share|cite|improve this answer













          Here $P_1$, $P_2$, and $P_3$ each represent a polynomial (with integer coefficients), and equations (1,2,3) are equations of polynomials in $x$ (not just single numbers). For instance, $P_1(x)$ is the polynomial you obtain by doing polynomial long division to divide $P(x)-b$ by $x-a$ (there will be no remainder since $P(a)=b$).



          In particular, if you plug in any integer value for $x$, these polynomials will give integer outputs. That's why $P_1(c)$ is an integer.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 6 hours ago









          Eric Wofsey

          161k12186297




          161k12186297




















              up vote
              1
              down vote













              From the given condition the polynomial $P(x)$ can be factorized as follow



              • $P(a)-b=0 implies P(x)-b=P_1(x)cdot (x-a)$

              where $P_1(x)$ is a polynomial of degree $n-1$ (assuming $P(x)$ is of degree $n$).



              Since $P_1(c)neq 0$ is an integer we have $|P_1(c)|ge 1$.



              Note that $P_1(c)neq 0$ since



              • $P(x)-b=P_1(x)cdot (x-a)implies P(c)-b=a-b=P_1(c)cdot (c-a)neq 0$





              share|cite|improve this answer

























                up vote
                1
                down vote













                From the given condition the polynomial $P(x)$ can be factorized as follow



                • $P(a)-b=0 implies P(x)-b=P_1(x)cdot (x-a)$

                where $P_1(x)$ is a polynomial of degree $n-1$ (assuming $P(x)$ is of degree $n$).



                Since $P_1(c)neq 0$ is an integer we have $|P_1(c)|ge 1$.



                Note that $P_1(c)neq 0$ since



                • $P(x)-b=P_1(x)cdot (x-a)implies P(c)-b=a-b=P_1(c)cdot (c-a)neq 0$





                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  From the given condition the polynomial $P(x)$ can be factorized as follow



                  • $P(a)-b=0 implies P(x)-b=P_1(x)cdot (x-a)$

                  where $P_1(x)$ is a polynomial of degree $n-1$ (assuming $P(x)$ is of degree $n$).



                  Since $P_1(c)neq 0$ is an integer we have $|P_1(c)|ge 1$.



                  Note that $P_1(c)neq 0$ since



                  • $P(x)-b=P_1(x)cdot (x-a)implies P(c)-b=a-b=P_1(c)cdot (c-a)neq 0$





                  share|cite|improve this answer













                  From the given condition the polynomial $P(x)$ can be factorized as follow



                  • $P(a)-b=0 implies P(x)-b=P_1(x)cdot (x-a)$

                  where $P_1(x)$ is a polynomial of degree $n-1$ (assuming $P(x)$ is of degree $n$).



                  Since $P_1(c)neq 0$ is an integer we have $|P_1(c)|ge 1$.



                  Note that $P_1(c)neq 0$ since



                  • $P(x)-b=P_1(x)cdot (x-a)implies P(c)-b=a-b=P_1(c)cdot (c-a)neq 0$






                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered 6 hours ago









                  gimusi

                  63.4k73379




                  63.4k73379






















                       

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