Am I the only one constantly forgetting the Eisenstein criterion? [on hold]

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Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.










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put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Maybe memorizing a few specific examples (and non-examples) can help.
    – Berci
    2 days ago






  • 1




    Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
    – André 3000
    2 days ago






  • 2




    I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
    – rschwieb
    2 days ago







  • 1




    How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
    – Jyrki Lahtonen
    2 days ago














up vote
5
down vote

favorite
5












Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.










share|cite|improve this question













put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Maybe memorizing a few specific examples (and non-examples) can help.
    – Berci
    2 days ago






  • 1




    Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
    – André 3000
    2 days ago






  • 2




    I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
    – rschwieb
    2 days ago







  • 1




    How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
    – Jyrki Lahtonen
    2 days ago












up vote
5
down vote

favorite
5









up vote
5
down vote

favorite
5






5





Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.










share|cite|improve this question













Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.







abstract-algebra intuition mnemonic






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asked 2 days ago









roi_saumon

1547




1547




put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    Maybe memorizing a few specific examples (and non-examples) can help.
    – Berci
    2 days ago






  • 1




    Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
    – André 3000
    2 days ago






  • 2




    I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
    – rschwieb
    2 days ago







  • 1




    How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
    – Jyrki Lahtonen
    2 days ago












  • 1




    Maybe memorizing a few specific examples (and non-examples) can help.
    – Berci
    2 days ago






  • 1




    Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
    – André 3000
    2 days ago






  • 2




    I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
    – rschwieb
    2 days ago







  • 1




    How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
    – Jyrki Lahtonen
    2 days ago







1




1




Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago




Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago




1




1




Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago




Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago




2




2




I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago





I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago





1




1




How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago




How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago










2 Answers
2






active

oldest

votes

















up vote
8
down vote



accepted










It's memorable if you view the proof as a consequence of the unique factorization of prime products.



The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$



By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$



But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.



Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$



Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.






share|cite|improve this answer






















  • Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
    – AOrtiz
    2 days ago










  • @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
    – Bill Dubuque
    2 days ago











  • Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
    – Bill Dubuque
    2 days ago











  • @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
    – roi_saumon
    12 hours ago










  • @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
    – Bill Dubuque
    11 hours ago

















up vote
6
down vote













The way I always remember it is via the Newton polygon.



The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.



Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.






share|cite|improve this answer



























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    8
    down vote



    accepted










    It's memorable if you view the proof as a consequence of the unique factorization of prime products.



    The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$



    By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$



    But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.



    Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$



    Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.






    share|cite|improve this answer






















    • Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
      – AOrtiz
      2 days ago










    • @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
      – Bill Dubuque
      2 days ago











    • Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
      – Bill Dubuque
      2 days ago











    • @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
      – roi_saumon
      12 hours ago










    • @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
      – Bill Dubuque
      11 hours ago














    up vote
    8
    down vote



    accepted










    It's memorable if you view the proof as a consequence of the unique factorization of prime products.



    The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$



    By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$



    But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.



    Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$



    Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.






    share|cite|improve this answer






















    • Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
      – AOrtiz
      2 days ago










    • @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
      – Bill Dubuque
      2 days ago











    • Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
      – Bill Dubuque
      2 days ago











    • @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
      – roi_saumon
      12 hours ago










    • @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
      – Bill Dubuque
      11 hours ago












    up vote
    8
    down vote



    accepted







    up vote
    8
    down vote



    accepted






    It's memorable if you view the proof as a consequence of the unique factorization of prime products.



    The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$



    By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$



    But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.



    Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$



    Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.






    share|cite|improve this answer














    It's memorable if you view the proof as a consequence of the unique factorization of prime products.



    The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$



    By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$



    But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.



    Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$



    Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    Bill Dubuque

    206k29189619




    206k29189619











    • Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
      – AOrtiz
      2 days ago










    • @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
      – Bill Dubuque
      2 days ago











    • Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
      – Bill Dubuque
      2 days ago











    • @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
      – roi_saumon
      12 hours ago










    • @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
      – Bill Dubuque
      11 hours ago
















    • Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
      – AOrtiz
      2 days ago










    • @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
      – Bill Dubuque
      2 days ago











    • Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
      – Bill Dubuque
      2 days ago











    • @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
      – roi_saumon
      12 hours ago










    • @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
      – Bill Dubuque
      11 hours ago















    Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
    – AOrtiz
    2 days ago




    Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
    – AOrtiz
    2 days ago












    @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
    – Bill Dubuque
    2 days ago





    @AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
    – Bill Dubuque
    2 days ago













    Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
    – Bill Dubuque
    2 days ago





    Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
    – Bill Dubuque
    2 days ago













    @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
    – roi_saumon
    12 hours ago




    @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
    – roi_saumon
    12 hours ago












    @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
    – Bill Dubuque
    11 hours ago




    @roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
    – Bill Dubuque
    11 hours ago










    up vote
    6
    down vote













    The way I always remember it is via the Newton polygon.



    The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.



    Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.






    share|cite|improve this answer
























      up vote
      6
      down vote













      The way I always remember it is via the Newton polygon.



      The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.



      Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.






      share|cite|improve this answer






















        up vote
        6
        down vote










        up vote
        6
        down vote









        The way I always remember it is via the Newton polygon.



        The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.



        Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.






        share|cite|improve this answer












        The way I always remember it is via the Newton polygon.



        The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.



        Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Alex J Best

        1,65711122




        1,65711122












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