Showing a polynomial having at least one integer root under certain conditions has precisely one integer root












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$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients. If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.



I tried by assuming a fifth degree polynomial but got stuck after that.



The question was asked by my friend.






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    7














    $P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients. If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.



    I tried by assuming a fifth degree polynomial but got stuck after that.



    The question was asked by my friend.






    polynomials







    share|cite|improve this question



























      7












      7








      7


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      $P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients. If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.



      I tried by assuming a fifth degree polynomial but got stuck after that.



      The question was asked by my friend.






      polynomials







      share|cite|improve this question















      $P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients. If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.



      I tried by assuming a fifth degree polynomial but got stuck after that.



      The question was asked by my friend.






      polynomials




      polynomials






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      share|cite|improve this question













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      edited 1 hour ago









      Eevee Trainer

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      asked 2 hours ago









      Ramanujam Ganit Prashikshan Ke

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          2 Answers
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          The assumption that $P$ has degree $5$ is irrelevant and unhelpful.



          If $r$ is a root of $P$, we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$, $P(a)=(a-r)Q(a)$ must be divisible by $a-r$. Taking $a=2$ and $a=10$, we see easily that the only possible value of $r$ is $-1$.



          Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$.






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            3














            If $u$ and $v$ are integer roots of $P$, then $P(x)=(x-u)(x-v)Q(x)$, where $Q$ is a polynomial with integer coefficients. From $P(2)=3$ we get $(u-2)(v-2)mid 3$, and then WLOG either $u-2=1$ or $u-2=-1$, implying $uin{1,3}$. Now $P(10)=11$ gives $(u-10)(v-10)mid 11$, showing that $u-10$ is a divisor of $11$. However, neither $1-10=-9$, not $3-10=-7$ is a divisor of $11$, a contradiction.






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              4














              The assumption that $P$ has degree $5$ is irrelevant and unhelpful.



              If $r$ is a root of $P$, we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$, $P(a)=(a-r)Q(a)$ must be divisible by $a-r$. Taking $a=2$ and $a=10$, we see easily that the only possible value of $r$ is $-1$.



              Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$.






              share|cite|improve this answer


























                4














                The assumption that $P$ has degree $5$ is irrelevant and unhelpful.



                If $r$ is a root of $P$, we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$, $P(a)=(a-r)Q(a)$ must be divisible by $a-r$. Taking $a=2$ and $a=10$, we see easily that the only possible value of $r$ is $-1$.



                Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$.






                share|cite|improve this answer
























                  4












                  4








                  4






                  The assumption that $P$ has degree $5$ is irrelevant and unhelpful.



                  If $r$ is a root of $P$, we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$, $P(a)=(a-r)Q(a)$ must be divisible by $a-r$. Taking $a=2$ and $a=10$, we see easily that the only possible value of $r$ is $-1$.



                  Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$.






                  share|cite|improve this answer












                  The assumption that $P$ has degree $5$ is irrelevant and unhelpful.



                  If $r$ is a root of $P$, we can write $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If $r$ is an integer, then $Q$ will also have integer coefficients (polynomial division never requires dividing coefficients if you are dividing by a monic polynomial). So, for any integer $a$, $P(a)=(a-r)Q(a)$ must be divisible by $a-r$. Taking $a=2$ and $a=10$, we see easily that the only possible value of $r$ is $-1$.



                  Moreover, we can say that $P$ only has one integer root even counting multiplicity, because if $-1$ were a root of higher multiplicity, we could write $P(x)=(x+1)^2R(x)$ where $R(x)$ again has integer coefficients, so $P(2)$ would need to be divisible by $(2+1)^2=9$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Eric Wofsey

                  179k12204331




                  179k12204331























                      3














                      If $u$ and $v$ are integer roots of $P$, then $P(x)=(x-u)(x-v)Q(x)$, where $Q$ is a polynomial with integer coefficients. From $P(2)=3$ we get $(u-2)(v-2)mid 3$, and then WLOG either $u-2=1$ or $u-2=-1$, implying $uin{1,3}$. Now $P(10)=11$ gives $(u-10)(v-10)mid 11$, showing that $u-10$ is a divisor of $11$. However, neither $1-10=-9$, not $3-10=-7$ is a divisor of $11$, a contradiction.






                      share|cite|improve this answer




























                        3














                        If $u$ and $v$ are integer roots of $P$, then $P(x)=(x-u)(x-v)Q(x)$, where $Q$ is a polynomial with integer coefficients. From $P(2)=3$ we get $(u-2)(v-2)mid 3$, and then WLOG either $u-2=1$ or $u-2=-1$, implying $uin{1,3}$. Now $P(10)=11$ gives $(u-10)(v-10)mid 11$, showing that $u-10$ is a divisor of $11$. However, neither $1-10=-9$, not $3-10=-7$ is a divisor of $11$, a contradiction.






                        share|cite|improve this answer


























                          3












                          3








                          3






                          If $u$ and $v$ are integer roots of $P$, then $P(x)=(x-u)(x-v)Q(x)$, where $Q$ is a polynomial with integer coefficients. From $P(2)=3$ we get $(u-2)(v-2)mid 3$, and then WLOG either $u-2=1$ or $u-2=-1$, implying $uin{1,3}$. Now $P(10)=11$ gives $(u-10)(v-10)mid 11$, showing that $u-10$ is a divisor of $11$. However, neither $1-10=-9$, not $3-10=-7$ is a divisor of $11$, a contradiction.






                          share|cite|improve this answer














                          If $u$ and $v$ are integer roots of $P$, then $P(x)=(x-u)(x-v)Q(x)$, where $Q$ is a polynomial with integer coefficients. From $P(2)=3$ we get $(u-2)(v-2)mid 3$, and then WLOG either $u-2=1$ or $u-2=-1$, implying $uin{1,3}$. Now $P(10)=11$ gives $(u-10)(v-10)mid 11$, showing that $u-10$ is a divisor of $11$. However, neither $1-10=-9$, not $3-10=-7$ is a divisor of $11$, a contradiction.







                          share|cite|improve this answer














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