symmetric polynomial recursion to solve the system, $x^5+y^5=33$, $x+y=3$

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I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 text , x+y=3$$



In the text they said to denote $sigma_1=x+y$ and $sigma_2=xy$, and to use recursion. I understand how the formula below works
$$S_5=x^5+y^5=sigma_1 S_4-sigma_2 S_3$$



But doing the recursion takes a long time. Is there a shorter way to do the recursion to solve this specific problem?



Thanks




algebra-precalculus polynomials systems-of-equations substitution symmetric-polynomials




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  • 1




    what is $x^2 + y^2 ; $ in terms of $sigma_2 ; ?$
    – Will Jagy
    Aug 6 at 1:07











  • Not sure I could get it just in terms of $sigma_2$, but I know that $x^2+y^2=(sigma_1)^2-2sigma_2$
    – john fowles
    Aug 6 at 1:11










  • and $sigma_1 = 3$
    – Will Jagy
    Aug 6 at 1:13






  • 1




    let me add an explicit hint, $xy^2 + x^2 y = xy(x+y)$
    – Will Jagy
    Aug 6 at 1:18






  • 1




    So the method is to consider the terms inside of $x^5$ and $y^5$ in the expansion of $(x+y)^5$? Since the polynomial is reciprocal, for the expansion of $(x^5+y^5)$ should I group those terms with the same coefficients. so I'd get $x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10xy(x^2+y^2)$
    – john fowles
    Aug 6 at 1:43














up vote
1
down vote

favorite












I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 text , x+y=3$$



In the text they said to denote $sigma_1=x+y$ and $sigma_2=xy$, and to use recursion. I understand how the formula below works
$$S_5=x^5+y^5=sigma_1 S_4-sigma_2 S_3$$



But doing the recursion takes a long time. Is there a shorter way to do the recursion to solve this specific problem?



Thanks




algebra-precalculus polynomials systems-of-equations substitution symmetric-polynomials




share|cite|improve this question

















  • 1




    what is $x^2 + y^2 ; $ in terms of $sigma_2 ; ?$
    – Will Jagy
    Aug 6 at 1:07











  • Not sure I could get it just in terms of $sigma_2$, but I know that $x^2+y^2=(sigma_1)^2-2sigma_2$
    – john fowles
    Aug 6 at 1:11










  • and $sigma_1 = 3$
    – Will Jagy
    Aug 6 at 1:13






  • 1




    let me add an explicit hint, $xy^2 + x^2 y = xy(x+y)$
    – Will Jagy
    Aug 6 at 1:18






  • 1




    So the method is to consider the terms inside of $x^5$ and $y^5$ in the expansion of $(x+y)^5$? Since the polynomial is reciprocal, for the expansion of $(x^5+y^5)$ should I group those terms with the same coefficients. so I'd get $x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10xy(x^2+y^2)$
    – john fowles
    Aug 6 at 1:43












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 text , x+y=3$$



In the text they said to denote $sigma_1=x+y$ and $sigma_2=xy$, and to use recursion. I understand how the formula below works
$$S_5=x^5+y^5=sigma_1 S_4-sigma_2 S_3$$



But doing the recursion takes a long time. Is there a shorter way to do the recursion to solve this specific problem?



Thanks




algebra-precalculus polynomials systems-of-equations substitution symmetric-polynomials




share|cite|improve this question













I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 text , x+y=3$$



In the text they said to denote $sigma_1=x+y$ and $sigma_2=xy$, and to use recursion. I understand how the formula below works
$$S_5=x^5+y^5=sigma_1 S_4-sigma_2 S_3$$



But doing the recursion takes a long time. Is there a shorter way to do the recursion to solve this specific problem?



Thanks





algebra-precalculus polynomials systems-of-equations substitution symmetric-polynomials





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 6:01









Michael Rozenberg

88.2k1579180




88.2k1579180









asked Aug 6 at 1:01









john fowles

1,093817




1,093817







  • 1




    what is $x^2 + y^2 ; $ in terms of $sigma_2 ; ?$
    – Will Jagy
    Aug 6 at 1:07











  • Not sure I could get it just in terms of $sigma_2$, but I know that $x^2+y^2=(sigma_1)^2-2sigma_2$
    – john fowles
    Aug 6 at 1:11










  • and $sigma_1 = 3$
    – Will Jagy
    Aug 6 at 1:13






  • 1




    let me add an explicit hint, $xy^2 + x^2 y = xy(x+y)$
    – Will Jagy
    Aug 6 at 1:18






  • 1




    So the method is to consider the terms inside of $x^5$ and $y^5$ in the expansion of $(x+y)^5$? Since the polynomial is reciprocal, for the expansion of $(x^5+y^5)$ should I group those terms with the same coefficients. so I'd get $x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10xy(x^2+y^2)$
    – john fowles
    Aug 6 at 1:43












  • 1




    what is $x^2 + y^2 ; $ in terms of $sigma_2 ; ?$
    – Will Jagy
    Aug 6 at 1:07











  • Not sure I could get it just in terms of $sigma_2$, but I know that $x^2+y^2=(sigma_1)^2-2sigma_2$
    – john fowles
    Aug 6 at 1:11










  • and $sigma_1 = 3$
    – Will Jagy
    Aug 6 at 1:13






  • 1




    let me add an explicit hint, $xy^2 + x^2 y = xy(x+y)$
    – Will Jagy
    Aug 6 at 1:18






  • 1




    So the method is to consider the terms inside of $x^5$ and $y^5$ in the expansion of $(x+y)^5$? Since the polynomial is reciprocal, for the expansion of $(x^5+y^5)$ should I group those terms with the same coefficients. so I'd get $x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10xy(x^2+y^2)$
    – john fowles
    Aug 6 at 1:43







1




1




what is $x^2 + y^2 ; $ in terms of $sigma_2 ; ?$
– Will Jagy
Aug 6 at 1:07





what is $x^2 + y^2 ; $ in terms of $sigma_2 ; ?$
– Will Jagy
Aug 6 at 1:07













Not sure I could get it just in terms of $sigma_2$, but I know that $x^2+y^2=(sigma_1)^2-2sigma_2$
– john fowles
Aug 6 at 1:11




Not sure I could get it just in terms of $sigma_2$, but I know that $x^2+y^2=(sigma_1)^2-2sigma_2$
– john fowles
Aug 6 at 1:11












and $sigma_1 = 3$
– Will Jagy
Aug 6 at 1:13




and $sigma_1 = 3$
– Will Jagy
Aug 6 at 1:13




1




1




let me add an explicit hint, $xy^2 + x^2 y = xy(x+y)$
– Will Jagy
Aug 6 at 1:18




let me add an explicit hint, $xy^2 + x^2 y = xy(x+y)$
– Will Jagy
Aug 6 at 1:18




1




1




So the method is to consider the terms inside of $x^5$ and $y^5$ in the expansion of $(x+y)^5$? Since the polynomial is reciprocal, for the expansion of $(x^5+y^5)$ should I group those terms with the same coefficients. so I'd get $x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10xy(x^2+y^2)$
– john fowles
Aug 6 at 1:43




So the method is to consider the terms inside of $x^5$ and $y^5$ in the expansion of $(x+y)^5$? Since the polynomial is reciprocal, for the expansion of $(x^5+y^5)$ should I group those terms with the same coefficients. so I'd get $x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10xy(x^2+y^2)$
– john fowles
Aug 6 at 1:43










3 Answers
3






active

oldest

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up vote
2
down vote













Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.



beginalign
x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
\
c&=
a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
\
c&=
x^4+y^4-b(x^2+y^2)+b^2
,\
x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
,\
c&=
a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
=
a(x^2+y^2)(x^2+y^2-b)-b^2)
,\
x^2+y^2&=(x+y)^2-2xy=a^2-2b
,\
c&=
a(a^2-2b)(a^2-2b-b)-ab^2
,
endalign



thus we have a quadratic equation in terms of $b$:



beginalign
5ab^2-5a^3b-c+a^5&=0
,
endalign



beginalign
b&=frac(5a^3pmsqrt5a^6+20ac10a
,\
b_1&=2
,\
b_2&=7
.
endalign



So, we need to consider two cases: $xy=2$ and $xy=7$.
In both cases we also have $x+y=3$.



Case 1 $xy=2$.



The quadratic equation with roots $x,y$ is
beginalign
t^2-3t+2&=0
,\
x,y&=1,2
.
endalign



Indeed, $1^5+2^5=33$, $1+2=3$,
so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.



Case 2 $xy=7$.



The quadratic equation with roots $x,y$ is
beginalign
t^2-3t+7&=0
,\
x,y&=tfrac32pmtfracsqrt192cdot i
,
endalign



so in Case 2 we have
two complex conjugate solutions,
beginalign
x&=tfrac32+tfracsqrt192cdot i
,\
y&=tfrac32-tfracsqrt192cdot i
endalign
and
beginalign
x&=tfrac32-tfracsqrt192cdot i
,\
y&=tfrac32+tfracsqrt192cdot i
.
endalign






share|cite|improve this answer





















  • But the OP wants a 'shorter way'...
    – Szeto
    Aug 6 at 3:05










  • @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
    – g.kov
    Aug 6 at 4:34

















up vote
1
down vote













Also, we can use the homogenization.



Let $y=tx$.



Thus, $$t^5+1=frac33(t+1)^5243$$ or
$$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.






share|cite|improve this answer




























    up vote
    0
    down vote














    Is there a shorter way to do the recursion to solve this specific problem?`




    If you don't insist on recursion, the shorter way in this case is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:



    $$
    beginalign
    0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \
    &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \
    &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7)
    endalign
    $$






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote













      Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.



      beginalign
      x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
      \
      c&=
      a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
      \
      c&=
      x^4+y^4-b(x^2+y^2)+b^2
      ,\
      x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
      ,\
      c&=
      a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
      =
      a(x^2+y^2)(x^2+y^2-b)-b^2)
      ,\
      x^2+y^2&=(x+y)^2-2xy=a^2-2b
      ,\
      c&=
      a(a^2-2b)(a^2-2b-b)-ab^2
      ,
      endalign



      thus we have a quadratic equation in terms of $b$:



      beginalign
      5ab^2-5a^3b-c+a^5&=0
      ,
      endalign



      beginalign
      b&=frac(5a^3pmsqrt5a^6+20ac10a
      ,\
      b_1&=2
      ,\
      b_2&=7
      .
      endalign



      So, we need to consider two cases: $xy=2$ and $xy=7$.
      In both cases we also have $x+y=3$.



      Case 1 $xy=2$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+2&=0
      ,\
      x,y&=1,2
      .
      endalign



      Indeed, $1^5+2^5=33$, $1+2=3$,
      so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.



      Case 2 $xy=7$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+7&=0
      ,\
      x,y&=tfrac32pmtfracsqrt192cdot i
      ,
      endalign



      so in Case 2 we have
      two complex conjugate solutions,
      beginalign
      x&=tfrac32+tfracsqrt192cdot i
      ,\
      y&=tfrac32-tfracsqrt192cdot i
      endalign
      and
      beginalign
      x&=tfrac32-tfracsqrt192cdot i
      ,\
      y&=tfrac32+tfracsqrt192cdot i
      .
      endalign






      share|cite|improve this answer





















      • But the OP wants a 'shorter way'...
        – Szeto
        Aug 6 at 3:05










      • @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
        – g.kov
        Aug 6 at 4:34














      up vote
      2
      down vote













      Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.



      beginalign
      x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
      \
      c&=
      a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
      \
      c&=
      x^4+y^4-b(x^2+y^2)+b^2
      ,\
      x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
      ,\
      c&=
      a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
      =
      a(x^2+y^2)(x^2+y^2-b)-b^2)
      ,\
      x^2+y^2&=(x+y)^2-2xy=a^2-2b
      ,\
      c&=
      a(a^2-2b)(a^2-2b-b)-ab^2
      ,
      endalign



      thus we have a quadratic equation in terms of $b$:



      beginalign
      5ab^2-5a^3b-c+a^5&=0
      ,
      endalign



      beginalign
      b&=frac(5a^3pmsqrt5a^6+20ac10a
      ,\
      b_1&=2
      ,\
      b_2&=7
      .
      endalign



      So, we need to consider two cases: $xy=2$ and $xy=7$.
      In both cases we also have $x+y=3$.



      Case 1 $xy=2$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+2&=0
      ,\
      x,y&=1,2
      .
      endalign



      Indeed, $1^5+2^5=33$, $1+2=3$,
      so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.



      Case 2 $xy=7$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+7&=0
      ,\
      x,y&=tfrac32pmtfracsqrt192cdot i
      ,
      endalign



      so in Case 2 we have
      two complex conjugate solutions,
      beginalign
      x&=tfrac32+tfracsqrt192cdot i
      ,\
      y&=tfrac32-tfracsqrt192cdot i
      endalign
      and
      beginalign
      x&=tfrac32-tfracsqrt192cdot i
      ,\
      y&=tfrac32+tfracsqrt192cdot i
      .
      endalign






      share|cite|improve this answer





















      • But the OP wants a 'shorter way'...
        – Szeto
        Aug 6 at 3:05










      • @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
        – g.kov
        Aug 6 at 4:34












      up vote
      2
      down vote










      up vote
      2
      down vote









      Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.



      beginalign
      x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
      \
      c&=
      a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
      \
      c&=
      x^4+y^4-b(x^2+y^2)+b^2
      ,\
      x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
      ,\
      c&=
      a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
      =
      a(x^2+y^2)(x^2+y^2-b)-b^2)
      ,\
      x^2+y^2&=(x+y)^2-2xy=a^2-2b
      ,\
      c&=
      a(a^2-2b)(a^2-2b-b)-ab^2
      ,
      endalign



      thus we have a quadratic equation in terms of $b$:



      beginalign
      5ab^2-5a^3b-c+a^5&=0
      ,
      endalign



      beginalign
      b&=frac(5a^3pmsqrt5a^6+20ac10a
      ,\
      b_1&=2
      ,\
      b_2&=7
      .
      endalign



      So, we need to consider two cases: $xy=2$ and $xy=7$.
      In both cases we also have $x+y=3$.



      Case 1 $xy=2$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+2&=0
      ,\
      x,y&=1,2
      .
      endalign



      Indeed, $1^5+2^5=33$, $1+2=3$,
      so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.



      Case 2 $xy=7$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+7&=0
      ,\
      x,y&=tfrac32pmtfracsqrt192cdot i
      ,
      endalign



      so in Case 2 we have
      two complex conjugate solutions,
      beginalign
      x&=tfrac32+tfracsqrt192cdot i
      ,\
      y&=tfrac32-tfracsqrt192cdot i
      endalign
      and
      beginalign
      x&=tfrac32-tfracsqrt192cdot i
      ,\
      y&=tfrac32+tfracsqrt192cdot i
      .
      endalign






      share|cite|improve this answer













      Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.



      beginalign
      x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
      \
      c&=
      a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
      \
      c&=
      x^4+y^4-b(x^2+y^2)+b^2
      ,\
      x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
      ,\
      c&=
      a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
      =
      a(x^2+y^2)(x^2+y^2-b)-b^2)
      ,\
      x^2+y^2&=(x+y)^2-2xy=a^2-2b
      ,\
      c&=
      a(a^2-2b)(a^2-2b-b)-ab^2
      ,
      endalign



      thus we have a quadratic equation in terms of $b$:



      beginalign
      5ab^2-5a^3b-c+a^5&=0
      ,
      endalign



      beginalign
      b&=frac(5a^3pmsqrt5a^6+20ac10a
      ,\
      b_1&=2
      ,\
      b_2&=7
      .
      endalign



      So, we need to consider two cases: $xy=2$ and $xy=7$.
      In both cases we also have $x+y=3$.



      Case 1 $xy=2$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+2&=0
      ,\
      x,y&=1,2
      .
      endalign



      Indeed, $1^5+2^5=33$, $1+2=3$,
      so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.



      Case 2 $xy=7$.



      The quadratic equation with roots $x,y$ is
      beginalign
      t^2-3t+7&=0
      ,\
      x,y&=tfrac32pmtfracsqrt192cdot i
      ,
      endalign



      so in Case 2 we have
      two complex conjugate solutions,
      beginalign
      x&=tfrac32+tfracsqrt192cdot i
      ,\
      y&=tfrac32-tfracsqrt192cdot i
      endalign
      and
      beginalign
      x&=tfrac32-tfracsqrt192cdot i
      ,\
      y&=tfrac32+tfracsqrt192cdot i
      .
      endalign







      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Aug 6 at 2:56









      g.kov

      5,5421717




      5,5421717











      • But the OP wants a 'shorter way'...
        – Szeto
        Aug 6 at 3:05










      • @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
        – g.kov
        Aug 6 at 4:34
















      • But the OP wants a 'shorter way'...
        – Szeto
        Aug 6 at 3:05










      • @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
        – g.kov
        Aug 6 at 4:34















      But the OP wants a 'shorter way'...
      – Szeto
      Aug 6 at 3:05




      But the OP wants a 'shorter way'...
      – Szeto
      Aug 6 at 3:05












      @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
      – g.kov
      Aug 6 at 4:34




      @Szeto: Well, technically, it can be considered somewhat shorter, since $S_3$ is not used.
      – g.kov
      Aug 6 at 4:34










      up vote
      1
      down vote













      Also, we can use the homogenization.



      Let $y=tx$.



      Thus, $$t^5+1=frac33(t+1)^5243$$ or
      $$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.






      share|cite|improve this answer

























        up vote
        1
        down vote













        Also, we can use the homogenization.



        Let $y=tx$.



        Thus, $$t^5+1=frac33(t+1)^5243$$ or
        $$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Also, we can use the homogenization.



          Let $y=tx$.



          Thus, $$t^5+1=frac33(t+1)^5243$$ or
          $$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.






          share|cite|improve this answer













          Also, we can use the homogenization.



          Let $y=tx$.



          Thus, $$t^5+1=frac33(t+1)^5243$$ or
          $$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 6 at 6:00









          Michael Rozenberg

          88.2k1579180




          88.2k1579180




















              up vote
              0
              down vote














              Is there a shorter way to do the recursion to solve this specific problem?`




              If you don't insist on recursion, the shorter way in this case is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:



              $$
              beginalign
              0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \
              &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \
              &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7)
              endalign
              $$






              share|cite|improve this answer

























                up vote
                0
                down vote














                Is there a shorter way to do the recursion to solve this specific problem?`




                If you don't insist on recursion, the shorter way in this case is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:



                $$
                beginalign
                0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \
                &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \
                &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7)
                endalign
                $$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote










                  Is there a shorter way to do the recursion to solve this specific problem?`




                  If you don't insist on recursion, the shorter way in this case is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:



                  $$
                  beginalign
                  0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \
                  &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \
                  &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7)
                  endalign
                  $$






                  share|cite|improve this answer














                  Is there a shorter way to do the recursion to solve this specific problem?`




                  If you don't insist on recursion, the shorter way in this case is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:



                  $$
                  beginalign
                  0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \
                  &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \
                  &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7)
                  endalign
                  $$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 7 at 2:48









                  dxiv

                  54.3k64797




                  54.3k64797






















                       

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