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Proof involving polynomial roots
Clash Royale CLAN TAG#URR8PPP
up vote
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From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.
polynomials roots
asked Jul 20 at 7:28
Merk Zockerborg
1429
add a comment |Â
up vote
7
down vote
favorite
From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.
polynomials roots
asked Jul 20 at 7:28
Merk Zockerborg
1429
1
You can see here artofproblemsolving.com/wiki/…
– Cesareo
Jul 20 at 7:42
By using your approach, how do you use the fact that $anot=b$?
– Robert Z
Jul 20 at 7:43
thanks, but I have already looked at that solution, and didn't really understand the motivation
– Merk Zockerborg
Jul 20 at 7:44
@RobertZ I subtracted the two equations from each other and factored out the $a-b$
– Merk Zockerborg
Jul 20 at 7:44
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.
polynomials roots
asked Jul 20 at 7:28
Merk Zockerborg
1429
From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.
polynomials roots
asked Jul 20 at 7:28
Merk Zockerborg
1429
asked Jul 20 at 7:28
Merk Zockerborg
1429
asked Jul 20 at 7:28
Merk Zockerborg
1429
asked Jul 20 at 7:28
Merk Zockerborg
1429
1429
1
You can see here artofproblemsolving.com/wiki/…
– Cesareo
Jul 20 at 7:42
By using your approach, how do you use the fact that $anot=b$?
– Robert Z
Jul 20 at 7:43
thanks, but I have already looked at that solution, and didn't really understand the motivation
– Merk Zockerborg
Jul 20 at 7:44
@RobertZ I subtracted the two equations from each other and factored out the $a-b$
– Merk Zockerborg
Jul 20 at 7:44
add a comment |Â
1
You can see here artofproblemsolving.com/wiki/…
– Cesareo
Jul 20 at 7:42
By using your approach, how do you use the fact that $anot=b$?
– Robert Z
Jul 20 at 7:43
thanks, but I have already looked at that solution, and didn't really understand the motivation
– Merk Zockerborg
Jul 20 at 7:44
@RobertZ I subtracted the two equations from each other and factored out the $a-b$
– Merk Zockerborg
Jul 20 at 7:44
1
1
You can see here artofproblemsolving.com/wiki/…
– Cesareo
Jul 20 at 7:42
You can see here artofproblemsolving.com/wiki/…
– Cesareo
Jul 20 at 7:42
By using your approach, how do you use the fact that $anot=b$?
– Robert Z
Jul 20 at 7:43
By using your approach, how do you use the fact that $anot=b$?
– Robert Z
Jul 20 at 7:43
thanks, but I have already looked at that solution, and didn't really understand the motivation
– Merk Zockerborg
Jul 20 at 7:44
thanks, but I have already looked at that solution, and didn't really understand the motivation
– Merk Zockerborg
Jul 20 at 7:44
@RobertZ I subtracted the two equations from each other and factored out the $a-b$
– Merk Zockerborg
Jul 20 at 7:44
@RobertZ I subtracted the two equations from each other and factored out the $a-b$
– Merk Zockerborg
Jul 20 at 7:44
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
Considering
beginalign
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \
S_1 &= a+b+c+d \
&= -1 \
S_2 &= ab+ac+ad+bc+bd+cd \
&= 0 \
S_3 &= bcd+acd+abd+abc \
&= 0 \
S_4 &= abcd \
&= -1 endalign
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),a,b,c,d] in Wolfram Alpha or so, we have:
beginalign
f(x) &= (x-colorredab)(x-colorredac)(x-colorredad)
(x-colorredbc)(x-colorredbd)(x-colorredcd) \
&= x^6-S_2 x^5+(S_1 S_3-S_4)x^4+(2S_2 S_4-S_3^2-S_1^2 S_4)x^3 \
& quad +(S_1 S_3 S_4-S_4^2)x^2-S_2 S_4^2 x+S_4^3 \
&= x^6+x^4+x^3-x^2-1
endalign
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
add a comment |Â
up vote
3
down vote
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $,ldots$
Following up on OP's approach, let $,ab=p,$ and, for symmetry, $,a+b=s,$. To begin with:
$$
begincases
beginalign
a^4 + a^3 - 1 &= 0 \
b^4 + b^3 - 1 &= 0
endalign
endcases tag1
$$
Subtracting the two equations in $,(1),$, and using that $,a ne b,$:
$$requirecancel
beginalign
a^4-b^4+a^3-b^3 = 0 ;;&implies;; cancel(a-b)(a+b)(a^2+b^2) + cancel(a-b)(a^2+ab+b^2) = 0 \
&implies;; s(s^2-2p) + s^2 - p = 0 \
&implies;; s^3 + s^2 - (2s+1)p = 0 tag2
endalign
$$
Adding the two equations in $,(1),$:
$$
beginalign
a^4+b^4+a^3+b^3 - 2 = 0 ;;&implies;; big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2big) \
&quadquadquadquad+big((a+b)^3-3ab(a+b)big)-2=0 \
&implies;; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \
&implies;; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 tag3
endalign
$$
At this point, the equation satisfied by $,p,$ can be derived by eliminating $,s,$ between $,(2),$ and $,(3),$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $,s^4+s^3=s cdot (2s+1)p,$ from $,(2),$ into $,(3),$:
$$
beginalign
s(2s+1)p -4ps^2-3ps+2p^2-2=0 ;;&implies;; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \
&implies;; ps^2+ps-p^2+1 = 0 tag4
endalign
$$
Eliminating once again $,s(s+1),$ between $,(2),$ and $,(4),$ by $,s cdot (4) - p cdot (2),$ gives:
$$
beginalign
0 &= s cdot big(cancelps(s+1)-p^2+1big) - p cdot big(cancels^2(s+1) - (2s+1)pbig) = (p^2+1)s+p^2 tag5
endalign
$$
Then, substituting $,s = frac-p^2p^2+1,$ back in $,(4),$ gives in the end the sextic $,p^6 + p^4 + p^3 - p^2 - 1 = 0,$.
answered Jul 21 at 1:36
dxiv
54.2k64797
add a comment |Â
up vote
0
down vote
I think that the problem is wrong as written, because if $a=b$, then $ab=a^2$ has minimal polynomial $x^4 - x^3 - 2 x^2 + 1$, which doesn't divide $x^6+x^4+x^3-x^2-1$ (the reminder is $7 x^3 + 6 x^2 - x - 5$).
I guess that it was meant that $aneq b$. Then, one may apply Galois theory to get a more straightforward way to compute the minimal polynomial of $ab$ and then to check whether or not it divides $x^6+x^4+x^3-x^2-1$ (it is a root iff the minimal polynomial divides...)
Here is a sketch how to do so: You prove that $mathbbQ(a,b)$ has degree $12$ and a basis $a^ib^j$ with $i=0,1,2,3$ and $j=0,1,2$. Then you write a matrix AB for multiplication by $ab$ in this basis (usually it is easier to write a matrix $A$ for $a$, a matrix $B$ for $b$, and to multiply them). The minimal polynomial polynomial will be that of the matrix $AB$.
The downside is that one needs to work with $12times 12$ matrices. But it is not too bad, since those matrices are essentially block-matrices.
answered Jul 20 at 15:30
Lior B-S
1,5661016
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Considering
beginalign
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \
S_1 &= a+b+c+d \
&= -1 \
S_2 &= ab+ac+ad+bc+bd+cd \
&= 0 \
S_3 &= bcd+acd+abd+abc \
&= 0 \
S_4 &= abcd \
&= -1 endalign
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),a,b,c,d] in Wolfram Alpha or so, we have:
beginalign
f(x) &= (x-colorredab)(x-colorredac)(x-colorredad)
(x-colorredbc)(x-colorredbd)(x-colorredcd) \
&= x^6-S_2 x^5+(S_1 S_3-S_4)x^4+(2S_2 S_4-S_3^2-S_1^2 S_4)x^3 \
& quad +(S_1 S_3 S_4-S_4^2)x^2-S_2 S_4^2 x+S_4^3 \
&= x^6+x^4+x^3-x^2-1
endalign
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
add a comment |Â
up vote
5
down vote
accepted
Considering
beginalign
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \
S_1 &= a+b+c+d \
&= -1 \
S_2 &= ab+ac+ad+bc+bd+cd \
&= 0 \
S_3 &= bcd+acd+abd+abc \
&= 0 \
S_4 &= abcd \
&= -1 endalign
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),a,b,c,d] in Wolfram Alpha or so, we have:
beginalign
f(x) &= (x-colorredab)(x-colorredac)(x-colorredad)
(x-colorredbc)(x-colorredbd)(x-colorredcd) \
&= x^6-S_2 x^5+(S_1 S_3-S_4)x^4+(2S_2 S_4-S_3^2-S_1^2 S_4)x^3 \
& quad +(S_1 S_3 S_4-S_4^2)x^2-S_2 S_4^2 x+S_4^3 \
&= x^6+x^4+x^3-x^2-1
endalign
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Considering
beginalign
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \
S_1 &= a+b+c+d \
&= -1 \
S_2 &= ab+ac+ad+bc+bd+cd \
&= 0 \
S_3 &= bcd+acd+abd+abc \
&= 0 \
S_4 &= abcd \
&= -1 endalign
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),a,b,c,d] in Wolfram Alpha or so, we have:
beginalign
f(x) &= (x-colorredab)(x-colorredac)(x-colorredad)
(x-colorredbc)(x-colorredbd)(x-colorredcd) \
&= x^6-S_2 x^5+(S_1 S_3-S_4)x^4+(2S_2 S_4-S_3^2-S_1^2 S_4)x^3 \
& quad +(S_1 S_3 S_4-S_4^2)x^2-S_2 S_4^2 x+S_4^3 \
&= x^6+x^4+x^3-x^2-1
endalign
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
Considering
beginalign
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \
S_1 &= a+b+c+d \
&= -1 \
S_2 &= ab+ac+ad+bc+bd+cd \
&= 0 \
S_3 &= bcd+acd+abd+abc \
&= 0 \
S_4 &= abcd \
&= -1 endalign
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),a,b,c,d] in Wolfram Alpha or so, we have:
beginalign
f(x) &= (x-colorredab)(x-colorredac)(x-colorredad)
(x-colorredbc)(x-colorredbd)(x-colorredcd) \
&= x^6-S_2 x^5+(S_1 S_3-S_4)x^4+(2S_2 S_4-S_3^2-S_1^2 S_4)x^3 \
& quad +(S_1 S_3 S_4-S_4^2)x^2-S_2 S_4^2 x+S_4^3 \
&= x^6+x^4+x^3-x^2-1
endalign
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
edited Jul 20 at 16:48
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
answered Jul 20 at 15:05
Ng Chung Tak
13k31130
13k31130
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
add a comment |Â
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
+1 Beautiful insight.
– Cave Johnson
Jul 21 at 1:42
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Can you explain why "The sextic has a nice symmetry so that all the roots are ab, ac, ad, bc, bd and cd" is true?
– Merk Zockerborg
Jul 21 at 16:36
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
Since $a$ and $b$ are not specified for which are which, all $a$, $b$, $c$ and $d$ have equal footings. That's to say any two of them will do. Hence there're $binom42=6$ combinations.
– Ng Chung Tak
Jul 21 at 19:24
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
I see, so you're working backwards, thanks
– Merk Zockerborg
Jul 22 at 14:59
add a comment |Â
up vote
3
down vote
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $,ldots$
Following up on OP's approach, let $,ab=p,$ and, for symmetry, $,a+b=s,$. To begin with:
$$
begincases
beginalign
a^4 + a^3 - 1 &= 0 \
b^4 + b^3 - 1 &= 0
endalign
endcases tag1
$$
Subtracting the two equations in $,(1),$, and using that $,a ne b,$:
$$requirecancel
beginalign
a^4-b^4+a^3-b^3 = 0 ;;&implies;; cancel(a-b)(a+b)(a^2+b^2) + cancel(a-b)(a^2+ab+b^2) = 0 \
&implies;; s(s^2-2p) + s^2 - p = 0 \
&implies;; s^3 + s^2 - (2s+1)p = 0 tag2
endalign
$$
Adding the two equations in $,(1),$:
$$
beginalign
a^4+b^4+a^3+b^3 - 2 = 0 ;;&implies;; big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2big) \
&quadquadquadquad+big((a+b)^3-3ab(a+b)big)-2=0 \
&implies;; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \
&implies;; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 tag3
endalign
$$
At this point, the equation satisfied by $,p,$ can be derived by eliminating $,s,$ between $,(2),$ and $,(3),$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $,s^4+s^3=s cdot (2s+1)p,$ from $,(2),$ into $,(3),$:
$$
beginalign
s(2s+1)p -4ps^2-3ps+2p^2-2=0 ;;&implies;; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \
&implies;; ps^2+ps-p^2+1 = 0 tag4
endalign
$$
Eliminating once again $,s(s+1),$ between $,(2),$ and $,(4),$ by $,s cdot (4) - p cdot (2),$ gives:
$$
beginalign
0 &= s cdot big(cancelps(s+1)-p^2+1big) - p cdot big(cancels^2(s+1) - (2s+1)pbig) = (p^2+1)s+p^2 tag5
endalign
$$
Then, substituting $,s = frac-p^2p^2+1,$ back in $,(4),$ gives in the end the sextic $,p^6 + p^4 + p^3 - p^2 - 1 = 0,$.
answered Jul 21 at 1:36
dxiv
54.2k64797
add a comment |Â
up vote
3
down vote
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $,ldots$
Following up on OP's approach, let $,ab=p,$ and, for symmetry, $,a+b=s,$. To begin with:
$$
begincases
beginalign
a^4 + a^3 - 1 &= 0 \
b^4 + b^3 - 1 &= 0
endalign
endcases tag1
$$
Subtracting the two equations in $,(1),$, and using that $,a ne b,$:
$$requirecancel
beginalign
a^4-b^4+a^3-b^3 = 0 ;;&implies;; cancel(a-b)(a+b)(a^2+b^2) + cancel(a-b)(a^2+ab+b^2) = 0 \
&implies;; s(s^2-2p) + s^2 - p = 0 \
&implies;; s^3 + s^2 - (2s+1)p = 0 tag2
endalign
$$
Adding the two equations in $,(1),$:
$$
beginalign
a^4+b^4+a^3+b^3 - 2 = 0 ;;&implies;; big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2big) \
&quadquadquadquad+big((a+b)^3-3ab(a+b)big)-2=0 \
&implies;; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \
&implies;; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 tag3
endalign
$$
At this point, the equation satisfied by $,p,$ can be derived by eliminating $,s,$ between $,(2),$ and $,(3),$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $,s^4+s^3=s cdot (2s+1)p,$ from $,(2),$ into $,(3),$:
$$
beginalign
s(2s+1)p -4ps^2-3ps+2p^2-2=0 ;;&implies;; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \
&implies;; ps^2+ps-p^2+1 = 0 tag4
endalign
$$
Eliminating once again $,s(s+1),$ between $,(2),$ and $,(4),$ by $,s cdot (4) - p cdot (2),$ gives:
$$
beginalign
0 &= s cdot big(cancelps(s+1)-p^2+1big) - p cdot big(cancels^2(s+1) - (2s+1)pbig) = (p^2+1)s+p^2 tag5
endalign
$$
Then, substituting $,s = frac-p^2p^2+1,$ back in $,(4),$ gives in the end the sextic $,p^6 + p^4 + p^3 - p^2 - 1 = 0,$.
answered Jul 21 at 1:36
dxiv
54.2k64797
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $,ldots$
Following up on OP's approach, let $,ab=p,$ and, for symmetry, $,a+b=s,$. To begin with:
$$
begincases
beginalign
a^4 + a^3 - 1 &= 0 \
b^4 + b^3 - 1 &= 0
endalign
endcases tag1
$$
Subtracting the two equations in $,(1),$, and using that $,a ne b,$:
$$requirecancel
beginalign
a^4-b^4+a^3-b^3 = 0 ;;&implies;; cancel(a-b)(a+b)(a^2+b^2) + cancel(a-b)(a^2+ab+b^2) = 0 \
&implies;; s(s^2-2p) + s^2 - p = 0 \
&implies;; s^3 + s^2 - (2s+1)p = 0 tag2
endalign
$$
Adding the two equations in $,(1),$:
$$
beginalign
a^4+b^4+a^3+b^3 - 2 = 0 ;;&implies;; big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2big) \
&quadquadquadquad+big((a+b)^3-3ab(a+b)big)-2=0 \
&implies;; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \
&implies;; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 tag3
endalign
$$
At this point, the equation satisfied by $,p,$ can be derived by eliminating $,s,$ between $,(2),$ and $,(3),$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $,s^4+s^3=s cdot (2s+1)p,$ from $,(2),$ into $,(3),$:
$$
beginalign
s(2s+1)p -4ps^2-3ps+2p^2-2=0 ;;&implies;; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \
&implies;; ps^2+ps-p^2+1 = 0 tag4
endalign
$$
Eliminating once again $,s(s+1),$ between $,(2),$ and $,(4),$ by $,s cdot (4) - p cdot (2),$ gives:
$$
beginalign
0 &= s cdot big(cancelps(s+1)-p^2+1big) - p cdot big(cancels^2(s+1) - (2s+1)pbig) = (p^2+1)s+p^2 tag5
endalign
$$
Then, substituting $,s = frac-p^2p^2+1,$ back in $,(4),$ gives in the end the sextic $,p^6 + p^4 + p^3 - p^2 - 1 = 0,$.
answered Jul 21 at 1:36
dxiv
54.2k64797
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $,ldots$
Following up on OP's approach, let $,ab=p,$ and, for symmetry, $,a+b=s,$. To begin with:
$$
begincases
beginalign
a^4 + a^3 - 1 &= 0 \
b^4 + b^3 - 1 &= 0
endalign
endcases tag1
$$
Subtracting the two equations in $,(1),$, and using that $,a ne b,$:
$$requirecancel
beginalign
a^4-b^4+a^3-b^3 = 0 ;;&implies;; cancel(a-b)(a+b)(a^2+b^2) + cancel(a-b)(a^2+ab+b^2) = 0 \
&implies;; s(s^2-2p) + s^2 - p = 0 \
&implies;; s^3 + s^2 - (2s+1)p = 0 tag2
endalign
$$
Adding the two equations in $,(1),$:
$$
beginalign
a^4+b^4+a^3+b^3 - 2 = 0 ;;&implies;; big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2big) \
&quadquadquadquad+big((a+b)^3-3ab(a+b)big)-2=0 \
&implies;; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \
&implies;; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 tag3
endalign
$$
At this point, the equation satisfied by $,p,$ can be derived by eliminating $,s,$ between $,(2),$ and $,(3),$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $,s^4+s^3=s cdot (2s+1)p,$ from $,(2),$ into $,(3),$:
$$
beginalign
s(2s+1)p -4ps^2-3ps+2p^2-2=0 ;;&implies;; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \
&implies;; ps^2+ps-p^2+1 = 0 tag4
endalign
$$
Eliminating once again $,s(s+1),$ between $,(2),$ and $,(4),$ by $,s cdot (4) - p cdot (2),$ gives:
$$
beginalign
0 &= s cdot big(cancelps(s+1)-p^2+1big) - p cdot big(cancels^2(s+1) - (2s+1)pbig) = (p^2+1)s+p^2 tag5
endalign
$$
Then, substituting $,s = frac-p^2p^2+1,$ back in $,(4),$ gives in the end the sextic $,p^6 + p^4 + p^3 - p^2 - 1 = 0,$.
answered Jul 21 at 1:36
dxiv
54.2k64797
answered Jul 21 at 1:36
dxiv
54.2k64797
answered Jul 21 at 1:36
dxiv
54.2k64797
answered Jul 21 at 1:36
dxiv
54.2k64797
54.2k64797
add a comment |Â
add a comment |Â
up vote
0
down vote
I think that the problem is wrong as written, because if $a=b$, then $ab=a^2$ has minimal polynomial $x^4 - x^3 - 2 x^2 + 1$, which doesn't divide $x^6+x^4+x^3-x^2-1$ (the reminder is $7 x^3 + 6 x^2 - x - 5$).
I guess that it was meant that $aneq b$. Then, one may apply Galois theory to get a more straightforward way to compute the minimal polynomial of $ab$ and then to check whether or not it divides $x^6+x^4+x^3-x^2-1$ (it is a root iff the minimal polynomial divides...)
Here is a sketch how to do so: You prove that $mathbbQ(a,b)$ has degree $12$ and a basis $a^ib^j$ with $i=0,1,2,3$ and $j=0,1,2$. Then you write a matrix AB for multiplication by $ab$ in this basis (usually it is easier to write a matrix $A$ for $a$, a matrix $B$ for $b$, and to multiply them). The minimal polynomial polynomial will be that of the matrix $AB$.
The downside is that one needs to work with $12times 12$ matrices. But it is not too bad, since those matrices are essentially block-matrices.
answered Jul 20 at 15:30
Lior B-S
1,5661016
add a comment |Â
up vote
0
down vote
I think that the problem is wrong as written, because if $a=b$, then $ab=a^2$ has minimal polynomial $x^4 - x^3 - 2 x^2 + 1$, which doesn't divide $x^6+x^4+x^3-x^2-1$ (the reminder is $7 x^3 + 6 x^2 - x - 5$).
I guess that it was meant that $aneq b$. Then, one may apply Galois theory to get a more straightforward way to compute the minimal polynomial of $ab$ and then to check whether or not it divides $x^6+x^4+x^3-x^2-1$ (it is a root iff the minimal polynomial divides...)
Here is a sketch how to do so: You prove that $mathbbQ(a,b)$ has degree $12$ and a basis $a^ib^j$ with $i=0,1,2,3$ and $j=0,1,2$. Then you write a matrix AB for multiplication by $ab$ in this basis (usually it is easier to write a matrix $A$ for $a$, a matrix $B$ for $b$, and to multiply them). The minimal polynomial polynomial will be that of the matrix $AB$.
The downside is that one needs to work with $12times 12$ matrices. But it is not too bad, since those matrices are essentially block-matrices.
answered Jul 20 at 15:30
Lior B-S
1,5661016
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think that the problem is wrong as written, because if $a=b$, then $ab=a^2$ has minimal polynomial $x^4 - x^3 - 2 x^2 + 1$, which doesn't divide $x^6+x^4+x^3-x^2-1$ (the reminder is $7 x^3 + 6 x^2 - x - 5$).
I guess that it was meant that $aneq b$. Then, one may apply Galois theory to get a more straightforward way to compute the minimal polynomial of $ab$ and then to check whether or not it divides $x^6+x^4+x^3-x^2-1$ (it is a root iff the minimal polynomial divides...)
Here is a sketch how to do so: You prove that $mathbbQ(a,b)$ has degree $12$ and a basis $a^ib^j$ with $i=0,1,2,3$ and $j=0,1,2$. Then you write a matrix AB for multiplication by $ab$ in this basis (usually it is easier to write a matrix $A$ for $a$, a matrix $B$ for $b$, and to multiply them). The minimal polynomial polynomial will be that of the matrix $AB$.
The downside is that one needs to work with $12times 12$ matrices. But it is not too bad, since those matrices are essentially block-matrices.
answered Jul 20 at 15:30
Lior B-S
1,5661016
I think that the problem is wrong as written, because if $a=b$, then $ab=a^2$ has minimal polynomial $x^4 - x^3 - 2 x^2 + 1$, which doesn't divide $x^6+x^4+x^3-x^2-1$ (the reminder is $7 x^3 + 6 x^2 - x - 5$).
I guess that it was meant that $aneq b$. Then, one may apply Galois theory to get a more straightforward way to compute the minimal polynomial of $ab$ and then to check whether or not it divides $x^6+x^4+x^3-x^2-1$ (it is a root iff the minimal polynomial divides...)
Here is a sketch how to do so: You prove that $mathbbQ(a,b)$ has degree $12$ and a basis $a^ib^j$ with $i=0,1,2,3$ and $j=0,1,2$. Then you write a matrix AB for multiplication by $ab$ in this basis (usually it is easier to write a matrix $A$ for $a$, a matrix $B$ for $b$, and to multiply them). The minimal polynomial polynomial will be that of the matrix $AB$.
The downside is that one needs to work with $12times 12$ matrices. But it is not too bad, since those matrices are essentially block-matrices.
answered Jul 20 at 15:30
Lior B-S
1,5661016
answered Jul 20 at 15:30
Lior B-S
1,5661016
answered Jul 20 at 15:30
Lior B-S
1,5661016
answered Jul 20 at 15:30
Lior B-S
1,5661016
1,5661016
add a comment |Â
add a comment |Â
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1
You can see here artofproblemsolving.com/wiki/…
– Cesareo
Jul 20 at 7:42
By using your approach, how do you use the fact that $anot=b$?
– Robert Z
Jul 20 at 7:43
thanks, but I have already looked at that solution, and didn't really understand the motivation
– Merk Zockerborg
Jul 20 at 7:44
@RobertZ I subtracted the two equations from each other and factored out the $a-b$
– Merk Zockerborg
Jul 20 at 7:44