Mixing
If $sqrt x + sqrt y + sqrt z = sqrt a + sqrt b + sqrt c$ is there a polynomial P such that $P(x, y, z, a, b,...
$begingroup$
If $sqrt x + sqrt y + sqrt z = sqrt a + sqrt b + sqrt c$ can we find a multivariable polynomial P such that $P(x, y, z, a, b, c) = 0$ ?
For an expression with less than six radicals, I know this is possible.
polynomials radicals
asked Jan 15 at 12:06
SmdSmd
886
$endgroup$
add a comment |
$begingroup$
If $sqrt x + sqrt y + sqrt z = sqrt a + sqrt b + sqrt c$ can we find a multivariable polynomial P such that $P(x, y, z, a, b, c) = 0$ ?
For an expression with less than six radicals, I know this is possible.
polynomials radicals
asked Jan 15 at 12:06
SmdSmd
886
$endgroup$
2
$begingroup$
Yes, for general reasons. Consider the polynomial ring $mathbb{Q}left[X,Y,Z,A,Bright]$, and set $C = X+Y+Z-A-B$ (so that $X+Y+Z=A+B+C$). Then, $X^2, Y^2, Z^2, A^2, B^2, C^2$ are $6$ elements of this ring, and thus are algebraically dependent (by Corollary 3 from mathoverflow.net/a/144625 ).
$endgroup$
– darij grinberg
Jan 15 at 12:43
1
$begingroup$
Yes, but it's big. With Mathematica, you can hit $sqrt{x}+sqrt{y}+sqrt{z}-sqrt{a}-sqrt{b}-sqrt{c}$ repeatedly with theResultantfunction to eliminate a square root at a time. The final result(ant) is a $16$-th degree polynomial with over $29,800$ terms.
$endgroup$
– Blue
Jan 15 at 13:04
add a comment |
$begingroup$
If $sqrt x + sqrt y + sqrt z = sqrt a + sqrt b + sqrt c$ can we find a multivariable polynomial P such that $P(x, y, z, a, b, c) = 0$ ?
For an expression with less than six radicals, I know this is possible.
polynomials radicals
asked Jan 15 at 12:06
SmdSmd
886
$endgroup$
If $sqrt x + sqrt y + sqrt z = sqrt a + sqrt b + sqrt c$ can we find a multivariable polynomial P such that $P(x, y, z, a, b, c) = 0$ ?
For an expression with less than six radicals, I know this is possible.
polynomials radicals
polynomials radicals
asked Jan 15 at 12:06
SmdSmd
886
asked Jan 15 at 12:06
SmdSmd
886
edited Jan 15 at 12:43
Smd
asked Jan 15 at 12:06
SmdSmd
886
asked Jan 15 at 12:06
SmdSmd
886
asked Jan 15 at 12:06
SmdSmd
886
886
2
$begingroup$
Yes, for general reasons. Consider the polynomial ring $mathbb{Q}left[X,Y,Z,A,Bright]$, and set $C = X+Y+Z-A-B$ (so that $X+Y+Z=A+B+C$). Then, $X^2, Y^2, Z^2, A^2, B^2, C^2$ are $6$ elements of this ring, and thus are algebraically dependent (by Corollary 3 from mathoverflow.net/a/144625 ).
$endgroup$
– darij grinberg
Jan 15 at 12:43
1
$begingroup$
Yes, but it's big. With Mathematica, you can hit $sqrt{x}+sqrt{y}+sqrt{z}-sqrt{a}-sqrt{b}-sqrt{c}$ repeatedly with theResultantfunction to eliminate a square root at a time. The final result(ant) is a $16$-th degree polynomial with over $29,800$ terms.
$endgroup$
– Blue
Jan 15 at 13:04
add a comment |
2
$begingroup$
Yes, for general reasons. Consider the polynomial ring $mathbb{Q}left[X,Y,Z,A,Bright]$, and set $C = X+Y+Z-A-B$ (so that $X+Y+Z=A+B+C$). Then, $X^2, Y^2, Z^2, A^2, B^2, C^2$ are $6$ elements of this ring, and thus are algebraically dependent (by Corollary 3 from mathoverflow.net/a/144625 ).
$endgroup$
– darij grinberg
Jan 15 at 12:43
1
$begingroup$
Yes, but it's big. With Mathematica, you can hit $sqrt{x}+sqrt{y}+sqrt{z}-sqrt{a}-sqrt{b}-sqrt{c}$ repeatedly with theResultantfunction to eliminate a square root at a time. The final result(ant) is a $16$-th degree polynomial with over $29,800$ terms.
$endgroup$
– Blue
Jan 15 at 13:04
2
2
$begingroup$
Yes, for general reasons. Consider the polynomial ring $mathbb{Q}left[X,Y,Z,A,Bright]$, and set $C = X+Y+Z-A-B$ (so that $X+Y+Z=A+B+C$). Then, $X^2, Y^2, Z^2, A^2, B^2, C^2$ are $6$ elements of this ring, and thus are algebraically dependent (by Corollary 3 from mathoverflow.net/a/144625 ).
$endgroup$
– darij grinberg
Jan 15 at 12:43
$begingroup$
Yes, for general reasons. Consider the polynomial ring $mathbb{Q}left[X,Y,Z,A,Bright]$, and set $C = X+Y+Z-A-B$ (so that $X+Y+Z=A+B+C$). Then, $X^2, Y^2, Z^2, A^2, B^2, C^2$ are $6$ elements of this ring, and thus are algebraically dependent (by Corollary 3 from mathoverflow.net/a/144625 ).
$endgroup$
– darij grinberg
Jan 15 at 12:43
1
1
$begingroup$
Yes, but it's big. With Mathematica, you can hit $sqrt{x}+sqrt{y}+sqrt{z}-sqrt{a}-sqrt{b}-sqrt{c}$ repeatedly with the
Resultant function to eliminate a square root at a time. The final result(ant) is a $16$-th degree polynomial with over $29,800$ terms.$endgroup$
– Blue
Jan 15 at 13:04
$begingroup$
Yes, but it's big. With Mathematica, you can hit $sqrt{x}+sqrt{y}+sqrt{z}-sqrt{a}-sqrt{b}-sqrt{c}$ repeatedly with the
Resultant function to eliminate a square root at a time. The final result(ant) is a $16$-th degree polynomial with over $29,800$ terms.$endgroup$
– Blue
Jan 15 at 13:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Expanding on a comment ...
With the help of Mathematica's Resultant function, we can systematically remove each square root from the expression. (As it turns out, we only need five applications of the function. The last square root goes away "for free".) This can, of course, be done for any number of radicals.
In this case, the final result(ant) is a $16$-th degree polynomial with over $29,800$ terms. For completeness, I'll give a representation of that polynomial below, using symmetric polynomials to condense the form:
$$begin{align}
s_1 &:= x + y + z + a + b + c \
s_2 &:= x y + x z + x a + x b + x c + cdots \
s_3 &:= x y z + x y a + x y b + x y c + cdots \
s_4 &:= x y z a + x y z b + x y z c + cdots \
s_5 &:= x y z a b + x y z a c + x y z b c + x y a b c + x z a b c + y z a b c \
s_6 &:= x y z a b c
end{align}$$
Here we go ...
$$begin{align}
P(x,y,z,a,b,c) &= s_1^{16} - 32 s_1^{14} s_2 + 448 s_1^{12} s_2^2 - 3584 s_1^{10} s_2^3 + 17920 s_1^8 s_2^4 - 57344 s_1^6 s_2^5 \
&+ 114688 s_1^4 s_2^6 - 131072 s_1^2 s_2^7 + 65536 s_2^8 - 256 s_1^{12} s_4 + 6144 s_1^{10} s_2 s_4 \
&- 61440 s_1^8 s_2^2 s_4 + 327680 s_1^6 s_2^3 s_4 - 983040 s_1^4 s_2^4 s_4 +
1572864 s_1^2 s_2^5 s_4 \
&- 1048576 s_2^6 s_4 + 24576 s_1^8 s_4^2 -
393216 s_1^6 s_2 s_4^2 + 2359296 s_1^4 s_2^2 s_4^2 \
&-
6291456 s_1^2 s_2^3 s_4^2 + 6291456 s_2^4 s_4^2 - 1048576 s_1^4 s_4^3 +
8388608 s_1^2 s_2 s_4^3 \
&- 16777216 s_2^2 s_4^3 + 16777216 s_4^4 -
4096 s_1^{11} s_5 + 81920 s_1^9 s_2 s_5 \
&- 655360 s_1^7 s_2^2 s_5 +
2621440 s_1^5 s_2^3 s_5 - 5242880 s_1^3 s_2^4 s_5 \
&+ 4194304 s_1 s_2^5 s_5 - 32768 s_1^8 s_3 s_5 + 524288 s_1^6 s_2 s_3 s_5 \
&- 3145728 s_1^4 s_2^2 s_3 s_5 + 8388608 s_1^2 s_2^3 s_3 s_5 - 8388608 s_2^4 s_3 s_5 \
&+ 524288 s_1^7 s_4 s_5 - 6291456 s_1^5 s_2 s_4 s_5 + 25165824 s_1^3 s_2^2 s_4 s_5 \
&- 33554432 s_1 s_2^3 s_4 s_5 + 4194304 s_1^4 s_3 s_4 s_5 -
33554432 s_1^2 s_2 s_3 s_4 s_5 \
&+ 67108864 s_2^2 s_3 s_4 s_5 - 16777216 s_1^3 s_4^2 s_5 + 67108864 s_1 s_2 s_4^2 s_5 \
&- 134217728 s_3 s_4^2 s_5 + 4194304 s_1^6 s_5^2 - 33554432 s_1^4 s_2 s_5^2 \
&+ 67108864 s_1^2 s_2^2 s_5^2 + 67108864 s_1^3 s_3 s_5^2 - 268435456 s_1 s_2 s_3 s_5^2 \
&+ 268435456 s_3^2 s_5^2 - 122880 s_1^{10} s_6 +
1900544 s_1^8 s_2 s_6 \
&- 10747904 s_1^6 s_2^2 s_6 + 25165824 s_1^4 s_2^3 s_6 -
14680064 s_1^2 s_2^4 s_6 \
&- 16777216 s_2^5 s_6 - 3145728 s_1^7 s_3 s_6 +
37748736 s_1^5 s_2 s_3 s_6 \
&- 150994944 s_1^3 s_2^2 s_3 s_6 + 201326592 s_1 s_2^3 s_3 s_6 - 16777216 s_1^4 s_3^2 s_6 \
&+ 134217728 s_1^2 s_2 s_3^2 s_6 - 268435456 s_2^2 s_3^2 s_6 -
9437184 s_1^6 s_4 s_6 \
&+ 83886080 s_1^4 s_2 s_4 s_6 - 218103808 s_1^2 s_2^2 s_4 s_6 + 134217728 s_2^3 s_4 s_6 \
&- 67108864 s_1^3 s_3 s_4 s_6 + 268435456 s_1 s_2 s_3 s_4 s_6 +
33554432 s_1^2 s_4^2 s_6 \
&- 268435456 s_2 s_4^2 s_6 - 25165824 s_1^5 s_5 s_6 + 134217728 s_1^3 s_2 s_5 s_6 \
&- 134217728 s_1 s_2^2 s_5 s_6 - 134217728 s_1^2 s_3 s_5 s_6 + 536870912 s_1 s_4 s_5 s_6 \
&- 1073741824 s_5^2 s_6 + 150994944 s_1^4 s_6^2 - 805306368 s_1^2 s_2 s_6^2 \
&+ 1073741824 s_2^2 s_6^2
end{align}$$
answered Jan 15 at 13:26
BlueBlue
48.5k870154
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add a comment |
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1 Answer
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1 Answer
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active
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active
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active
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votes
$begingroup$
Expanding on a comment ...
With the help of Mathematica's Resultant function, we can systematically remove each square root from the expression. (As it turns out, we only need five applications of the function. The last square root goes away "for free".) This can, of course, be done for any number of radicals.
In this case, the final result(ant) is a $16$-th degree polynomial with over $29,800$ terms. For completeness, I'll give a representation of that polynomial below, using symmetric polynomials to condense the form:
$$begin{align}
s_1 &:= x + y + z + a + b + c \
s_2 &:= x y + x z + x a + x b + x c + cdots \
s_3 &:= x y z + x y a + x y b + x y c + cdots \
s_4 &:= x y z a + x y z b + x y z c + cdots \
s_5 &:= x y z a b + x y z a c + x y z b c + x y a b c + x z a b c + y z a b c \
s_6 &:= x y z a b c
end{align}$$
Here we go ...
$$begin{align}
P(x,y,z,a,b,c) &= s_1^{16} - 32 s_1^{14} s_2 + 448 s_1^{12} s_2^2 - 3584 s_1^{10} s_2^3 + 17920 s_1^8 s_2^4 - 57344 s_1^6 s_2^5 \
&+ 114688 s_1^4 s_2^6 - 131072 s_1^2 s_2^7 + 65536 s_2^8 - 256 s_1^{12} s_4 + 6144 s_1^{10} s_2 s_4 \
&- 61440 s_1^8 s_2^2 s_4 + 327680 s_1^6 s_2^3 s_4 - 983040 s_1^4 s_2^4 s_4 +
1572864 s_1^2 s_2^5 s_4 \
&- 1048576 s_2^6 s_4 + 24576 s_1^8 s_4^2 -
393216 s_1^6 s_2 s_4^2 + 2359296 s_1^4 s_2^2 s_4^2 \
&-
6291456 s_1^2 s_2^3 s_4^2 + 6291456 s_2^4 s_4^2 - 1048576 s_1^4 s_4^3 +
8388608 s_1^2 s_2 s_4^3 \
&- 16777216 s_2^2 s_4^3 + 16777216 s_4^4 -
4096 s_1^{11} s_5 + 81920 s_1^9 s_2 s_5 \
&- 655360 s_1^7 s_2^2 s_5 +
2621440 s_1^5 s_2^3 s_5 - 5242880 s_1^3 s_2^4 s_5 \
&+ 4194304 s_1 s_2^5 s_5 - 32768 s_1^8 s_3 s_5 + 524288 s_1^6 s_2 s_3 s_5 \
&- 3145728 s_1^4 s_2^2 s_3 s_5 + 8388608 s_1^2 s_2^3 s_3 s_5 - 8388608 s_2^4 s_3 s_5 \
&+ 524288 s_1^7 s_4 s_5 - 6291456 s_1^5 s_2 s_4 s_5 + 25165824 s_1^3 s_2^2 s_4 s_5 \
&- 33554432 s_1 s_2^3 s_4 s_5 + 4194304 s_1^4 s_3 s_4 s_5 -
33554432 s_1^2 s_2 s_3 s_4 s_5 \
&+ 67108864 s_2^2 s_3 s_4 s_5 - 16777216 s_1^3 s_4^2 s_5 + 67108864 s_1 s_2 s_4^2 s_5 \
&- 134217728 s_3 s_4^2 s_5 + 4194304 s_1^6 s_5^2 - 33554432 s_1^4 s_2 s_5^2 \
&+ 67108864 s_1^2 s_2^2 s_5^2 + 67108864 s_1^3 s_3 s_5^2 - 268435456 s_1 s_2 s_3 s_5^2 \
&+ 268435456 s_3^2 s_5^2 - 122880 s_1^{10} s_6 +
1900544 s_1^8 s_2 s_6 \
&- 10747904 s_1^6 s_2^2 s_6 + 25165824 s_1^4 s_2^3 s_6 -
14680064 s_1^2 s_2^4 s_6 \
&- 16777216 s_2^5 s_6 - 3145728 s_1^7 s_3 s_6 +
37748736 s_1^5 s_2 s_3 s_6 \
&- 150994944 s_1^3 s_2^2 s_3 s_6 + 201326592 s_1 s_2^3 s_3 s_6 - 16777216 s_1^4 s_3^2 s_6 \
&+ 134217728 s_1^2 s_2 s_3^2 s_6 - 268435456 s_2^2 s_3^2 s_6 -
9437184 s_1^6 s_4 s_6 \
&+ 83886080 s_1^4 s_2 s_4 s_6 - 218103808 s_1^2 s_2^2 s_4 s_6 + 134217728 s_2^3 s_4 s_6 \
&- 67108864 s_1^3 s_3 s_4 s_6 + 268435456 s_1 s_2 s_3 s_4 s_6 +
33554432 s_1^2 s_4^2 s_6 \
&- 268435456 s_2 s_4^2 s_6 - 25165824 s_1^5 s_5 s_6 + 134217728 s_1^3 s_2 s_5 s_6 \
&- 134217728 s_1 s_2^2 s_5 s_6 - 134217728 s_1^2 s_3 s_5 s_6 + 536870912 s_1 s_4 s_5 s_6 \
&- 1073741824 s_5^2 s_6 + 150994944 s_1^4 s_6^2 - 805306368 s_1^2 s_2 s_6^2 \
&+ 1073741824 s_2^2 s_6^2
end{align}$$
answered Jan 15 at 13:26
BlueBlue
48.5k870154
$endgroup$
add a comment |
$begingroup$
Expanding on a comment ...
With the help of Mathematica's Resultant function, we can systematically remove each square root from the expression. (As it turns out, we only need five applications of the function. The last square root goes away "for free".) This can, of course, be done for any number of radicals.
In this case, the final result(ant) is a $16$-th degree polynomial with over $29,800$ terms. For completeness, I'll give a representation of that polynomial below, using symmetric polynomials to condense the form:
$$begin{align}
s_1 &:= x + y + z + a + b + c \
s_2 &:= x y + x z + x a + x b + x c + cdots \
s_3 &:= x y z + x y a + x y b + x y c + cdots \
s_4 &:= x y z a + x y z b + x y z c + cdots \
s_5 &:= x y z a b + x y z a c + x y z b c + x y a b c + x z a b c + y z a b c \
s_6 &:= x y z a b c
end{align}$$
Here we go ...
$$begin{align}
P(x,y,z,a,b,c) &= s_1^{16} - 32 s_1^{14} s_2 + 448 s_1^{12} s_2^2 - 3584 s_1^{10} s_2^3 + 17920 s_1^8 s_2^4 - 57344 s_1^6 s_2^5 \
&+ 114688 s_1^4 s_2^6 - 131072 s_1^2 s_2^7 + 65536 s_2^8 - 256 s_1^{12} s_4 + 6144 s_1^{10} s_2 s_4 \
&- 61440 s_1^8 s_2^2 s_4 + 327680 s_1^6 s_2^3 s_4 - 983040 s_1^4 s_2^4 s_4 +
1572864 s_1^2 s_2^5 s_4 \
&- 1048576 s_2^6 s_4 + 24576 s_1^8 s_4^2 -
393216 s_1^6 s_2 s_4^2 + 2359296 s_1^4 s_2^2 s_4^2 \
&-
6291456 s_1^2 s_2^3 s_4^2 + 6291456 s_2^4 s_4^2 - 1048576 s_1^4 s_4^3 +
8388608 s_1^2 s_2 s_4^3 \
&- 16777216 s_2^2 s_4^3 + 16777216 s_4^4 -
4096 s_1^{11} s_5 + 81920 s_1^9 s_2 s_5 \
&- 655360 s_1^7 s_2^2 s_5 +
2621440 s_1^5 s_2^3 s_5 - 5242880 s_1^3 s_2^4 s_5 \
&+ 4194304 s_1 s_2^5 s_5 - 32768 s_1^8 s_3 s_5 + 524288 s_1^6 s_2 s_3 s_5 \
&- 3145728 s_1^4 s_2^2 s_3 s_5 + 8388608 s_1^2 s_2^3 s_3 s_5 - 8388608 s_2^4 s_3 s_5 \
&+ 524288 s_1^7 s_4 s_5 - 6291456 s_1^5 s_2 s_4 s_5 + 25165824 s_1^3 s_2^2 s_4 s_5 \
&- 33554432 s_1 s_2^3 s_4 s_5 + 4194304 s_1^4 s_3 s_4 s_5 -
33554432 s_1^2 s_2 s_3 s_4 s_5 \
&+ 67108864 s_2^2 s_3 s_4 s_5 - 16777216 s_1^3 s_4^2 s_5 + 67108864 s_1 s_2 s_4^2 s_5 \
&- 134217728 s_3 s_4^2 s_5 + 4194304 s_1^6 s_5^2 - 33554432 s_1^4 s_2 s_5^2 \
&+ 67108864 s_1^2 s_2^2 s_5^2 + 67108864 s_1^3 s_3 s_5^2 - 268435456 s_1 s_2 s_3 s_5^2 \
&+ 268435456 s_3^2 s_5^2 - 122880 s_1^{10} s_6 +
1900544 s_1^8 s_2 s_6 \
&- 10747904 s_1^6 s_2^2 s_6 + 25165824 s_1^4 s_2^3 s_6 -
14680064 s_1^2 s_2^4 s_6 \
&- 16777216 s_2^5 s_6 - 3145728 s_1^7 s_3 s_6 +
37748736 s_1^5 s_2 s_3 s_6 \
&- 150994944 s_1^3 s_2^2 s_3 s_6 + 201326592 s_1 s_2^3 s_3 s_6 - 16777216 s_1^4 s_3^2 s_6 \
&+ 134217728 s_1^2 s_2 s_3^2 s_6 - 268435456 s_2^2 s_3^2 s_6 -
9437184 s_1^6 s_4 s_6 \
&+ 83886080 s_1^4 s_2 s_4 s_6 - 218103808 s_1^2 s_2^2 s_4 s_6 + 134217728 s_2^3 s_4 s_6 \
&- 67108864 s_1^3 s_3 s_4 s_6 + 268435456 s_1 s_2 s_3 s_4 s_6 +
33554432 s_1^2 s_4^2 s_6 \
&- 268435456 s_2 s_4^2 s_6 - 25165824 s_1^5 s_5 s_6 + 134217728 s_1^3 s_2 s_5 s_6 \
&- 134217728 s_1 s_2^2 s_5 s_6 - 134217728 s_1^2 s_3 s_5 s_6 + 536870912 s_1 s_4 s_5 s_6 \
&- 1073741824 s_5^2 s_6 + 150994944 s_1^4 s_6^2 - 805306368 s_1^2 s_2 s_6^2 \
&+ 1073741824 s_2^2 s_6^2
end{align}$$
answered Jan 15 at 13:26
BlueBlue
48.5k870154
$endgroup$
add a comment |
$begingroup$
Expanding on a comment ...
With the help of Mathematica's Resultant function, we can systematically remove each square root from the expression. (As it turns out, we only need five applications of the function. The last square root goes away "for free".) This can, of course, be done for any number of radicals.
In this case, the final result(ant) is a $16$-th degree polynomial with over $29,800$ terms. For completeness, I'll give a representation of that polynomial below, using symmetric polynomials to condense the form:
$$begin{align}
s_1 &:= x + y + z + a + b + c \
s_2 &:= x y + x z + x a + x b + x c + cdots \
s_3 &:= x y z + x y a + x y b + x y c + cdots \
s_4 &:= x y z a + x y z b + x y z c + cdots \
s_5 &:= x y z a b + x y z a c + x y z b c + x y a b c + x z a b c + y z a b c \
s_6 &:= x y z a b c
end{align}$$
Here we go ...
$$begin{align}
P(x,y,z,a,b,c) &= s_1^{16} - 32 s_1^{14} s_2 + 448 s_1^{12} s_2^2 - 3584 s_1^{10} s_2^3 + 17920 s_1^8 s_2^4 - 57344 s_1^6 s_2^5 \
&+ 114688 s_1^4 s_2^6 - 131072 s_1^2 s_2^7 + 65536 s_2^8 - 256 s_1^{12} s_4 + 6144 s_1^{10} s_2 s_4 \
&- 61440 s_1^8 s_2^2 s_4 + 327680 s_1^6 s_2^3 s_4 - 983040 s_1^4 s_2^4 s_4 +
1572864 s_1^2 s_2^5 s_4 \
&- 1048576 s_2^6 s_4 + 24576 s_1^8 s_4^2 -
393216 s_1^6 s_2 s_4^2 + 2359296 s_1^4 s_2^2 s_4^2 \
&-
6291456 s_1^2 s_2^3 s_4^2 + 6291456 s_2^4 s_4^2 - 1048576 s_1^4 s_4^3 +
8388608 s_1^2 s_2 s_4^3 \
&- 16777216 s_2^2 s_4^3 + 16777216 s_4^4 -
4096 s_1^{11} s_5 + 81920 s_1^9 s_2 s_5 \
&- 655360 s_1^7 s_2^2 s_5 +
2621440 s_1^5 s_2^3 s_5 - 5242880 s_1^3 s_2^4 s_5 \
&+ 4194304 s_1 s_2^5 s_5 - 32768 s_1^8 s_3 s_5 + 524288 s_1^6 s_2 s_3 s_5 \
&- 3145728 s_1^4 s_2^2 s_3 s_5 + 8388608 s_1^2 s_2^3 s_3 s_5 - 8388608 s_2^4 s_3 s_5 \
&+ 524288 s_1^7 s_4 s_5 - 6291456 s_1^5 s_2 s_4 s_5 + 25165824 s_1^3 s_2^2 s_4 s_5 \
&- 33554432 s_1 s_2^3 s_4 s_5 + 4194304 s_1^4 s_3 s_4 s_5 -
33554432 s_1^2 s_2 s_3 s_4 s_5 \
&+ 67108864 s_2^2 s_3 s_4 s_5 - 16777216 s_1^3 s_4^2 s_5 + 67108864 s_1 s_2 s_4^2 s_5 \
&- 134217728 s_3 s_4^2 s_5 + 4194304 s_1^6 s_5^2 - 33554432 s_1^4 s_2 s_5^2 \
&+ 67108864 s_1^2 s_2^2 s_5^2 + 67108864 s_1^3 s_3 s_5^2 - 268435456 s_1 s_2 s_3 s_5^2 \
&+ 268435456 s_3^2 s_5^2 - 122880 s_1^{10} s_6 +
1900544 s_1^8 s_2 s_6 \
&- 10747904 s_1^6 s_2^2 s_6 + 25165824 s_1^4 s_2^3 s_6 -
14680064 s_1^2 s_2^4 s_6 \
&- 16777216 s_2^5 s_6 - 3145728 s_1^7 s_3 s_6 +
37748736 s_1^5 s_2 s_3 s_6 \
&- 150994944 s_1^3 s_2^2 s_3 s_6 + 201326592 s_1 s_2^3 s_3 s_6 - 16777216 s_1^4 s_3^2 s_6 \
&+ 134217728 s_1^2 s_2 s_3^2 s_6 - 268435456 s_2^2 s_3^2 s_6 -
9437184 s_1^6 s_4 s_6 \
&+ 83886080 s_1^4 s_2 s_4 s_6 - 218103808 s_1^2 s_2^2 s_4 s_6 + 134217728 s_2^3 s_4 s_6 \
&- 67108864 s_1^3 s_3 s_4 s_6 + 268435456 s_1 s_2 s_3 s_4 s_6 +
33554432 s_1^2 s_4^2 s_6 \
&- 268435456 s_2 s_4^2 s_6 - 25165824 s_1^5 s_5 s_6 + 134217728 s_1^3 s_2 s_5 s_6 \
&- 134217728 s_1 s_2^2 s_5 s_6 - 134217728 s_1^2 s_3 s_5 s_6 + 536870912 s_1 s_4 s_5 s_6 \
&- 1073741824 s_5^2 s_6 + 150994944 s_1^4 s_6^2 - 805306368 s_1^2 s_2 s_6^2 \
&+ 1073741824 s_2^2 s_6^2
end{align}$$
answered Jan 15 at 13:26
BlueBlue
48.5k870154
$endgroup$
Expanding on a comment ...
With the help of Mathematica's Resultant function, we can systematically remove each square root from the expression. (As it turns out, we only need five applications of the function. The last square root goes away "for free".) This can, of course, be done for any number of radicals.
In this case, the final result(ant) is a $16$-th degree polynomial with over $29,800$ terms. For completeness, I'll give a representation of that polynomial below, using symmetric polynomials to condense the form:
$$begin{align}
s_1 &:= x + y + z + a + b + c \
s_2 &:= x y + x z + x a + x b + x c + cdots \
s_3 &:= x y z + x y a + x y b + x y c + cdots \
s_4 &:= x y z a + x y z b + x y z c + cdots \
s_5 &:= x y z a b + x y z a c + x y z b c + x y a b c + x z a b c + y z a b c \
s_6 &:= x y z a b c
end{align}$$
Here we go ...
$$begin{align}
P(x,y,z,a,b,c) &= s_1^{16} - 32 s_1^{14} s_2 + 448 s_1^{12} s_2^2 - 3584 s_1^{10} s_2^3 + 17920 s_1^8 s_2^4 - 57344 s_1^6 s_2^5 \
&+ 114688 s_1^4 s_2^6 - 131072 s_1^2 s_2^7 + 65536 s_2^8 - 256 s_1^{12} s_4 + 6144 s_1^{10} s_2 s_4 \
&- 61440 s_1^8 s_2^2 s_4 + 327680 s_1^6 s_2^3 s_4 - 983040 s_1^4 s_2^4 s_4 +
1572864 s_1^2 s_2^5 s_4 \
&- 1048576 s_2^6 s_4 + 24576 s_1^8 s_4^2 -
393216 s_1^6 s_2 s_4^2 + 2359296 s_1^4 s_2^2 s_4^2 \
&-
6291456 s_1^2 s_2^3 s_4^2 + 6291456 s_2^4 s_4^2 - 1048576 s_1^4 s_4^3 +
8388608 s_1^2 s_2 s_4^3 \
&- 16777216 s_2^2 s_4^3 + 16777216 s_4^4 -
4096 s_1^{11} s_5 + 81920 s_1^9 s_2 s_5 \
&- 655360 s_1^7 s_2^2 s_5 +
2621440 s_1^5 s_2^3 s_5 - 5242880 s_1^3 s_2^4 s_5 \
&+ 4194304 s_1 s_2^5 s_5 - 32768 s_1^8 s_3 s_5 + 524288 s_1^6 s_2 s_3 s_5 \
&- 3145728 s_1^4 s_2^2 s_3 s_5 + 8388608 s_1^2 s_2^3 s_3 s_5 - 8388608 s_2^4 s_3 s_5 \
&+ 524288 s_1^7 s_4 s_5 - 6291456 s_1^5 s_2 s_4 s_5 + 25165824 s_1^3 s_2^2 s_4 s_5 \
&- 33554432 s_1 s_2^3 s_4 s_5 + 4194304 s_1^4 s_3 s_4 s_5 -
33554432 s_1^2 s_2 s_3 s_4 s_5 \
&+ 67108864 s_2^2 s_3 s_4 s_5 - 16777216 s_1^3 s_4^2 s_5 + 67108864 s_1 s_2 s_4^2 s_5 \
&- 134217728 s_3 s_4^2 s_5 + 4194304 s_1^6 s_5^2 - 33554432 s_1^4 s_2 s_5^2 \
&+ 67108864 s_1^2 s_2^2 s_5^2 + 67108864 s_1^3 s_3 s_5^2 - 268435456 s_1 s_2 s_3 s_5^2 \
&+ 268435456 s_3^2 s_5^2 - 122880 s_1^{10} s_6 +
1900544 s_1^8 s_2 s_6 \
&- 10747904 s_1^6 s_2^2 s_6 + 25165824 s_1^4 s_2^3 s_6 -
14680064 s_1^2 s_2^4 s_6 \
&- 16777216 s_2^5 s_6 - 3145728 s_1^7 s_3 s_6 +
37748736 s_1^5 s_2 s_3 s_6 \
&- 150994944 s_1^3 s_2^2 s_3 s_6 + 201326592 s_1 s_2^3 s_3 s_6 - 16777216 s_1^4 s_3^2 s_6 \
&+ 134217728 s_1^2 s_2 s_3^2 s_6 - 268435456 s_2^2 s_3^2 s_6 -
9437184 s_1^6 s_4 s_6 \
&+ 83886080 s_1^4 s_2 s_4 s_6 - 218103808 s_1^2 s_2^2 s_4 s_6 + 134217728 s_2^3 s_4 s_6 \
&- 67108864 s_1^3 s_3 s_4 s_6 + 268435456 s_1 s_2 s_3 s_4 s_6 +
33554432 s_1^2 s_4^2 s_6 \
&- 268435456 s_2 s_4^2 s_6 - 25165824 s_1^5 s_5 s_6 + 134217728 s_1^3 s_2 s_5 s_6 \
&- 134217728 s_1 s_2^2 s_5 s_6 - 134217728 s_1^2 s_3 s_5 s_6 + 536870912 s_1 s_4 s_5 s_6 \
&- 1073741824 s_5^2 s_6 + 150994944 s_1^4 s_6^2 - 805306368 s_1^2 s_2 s_6^2 \
&+ 1073741824 s_2^2 s_6^2
end{align}$$
answered Jan 15 at 13:26
BlueBlue
48.5k870154
edited Jan 15 at 13:31
answered Jan 15 at 13:26
BlueBlue
48.5k870154
answered Jan 15 at 13:26
BlueBlue
48.5k870154
answered Jan 15 at 13:26
BlueBlue
48.5k870154
48.5k870154
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2
$begingroup$
Yes, for general reasons. Consider the polynomial ring $mathbb{Q}left[X,Y,Z,A,Bright]$, and set $C = X+Y+Z-A-B$ (so that $X+Y+Z=A+B+C$). Then, $X^2, Y^2, Z^2, A^2, B^2, C^2$ are $6$ elements of this ring, and thus are algebraically dependent (by Corollary 3 from mathoverflow.net/a/144625 ).
$endgroup$
– darij grinberg
Jan 15 at 12:43
1
$begingroup$
Yes, but it's big. With Mathematica, you can hit $sqrt{x}+sqrt{y}+sqrt{z}-sqrt{a}-sqrt{b}-sqrt{c}$ repeatedly with the
Resultantfunction to eliminate a square root at a time. The final result(ant) is a $16$-th degree polynomial with over $29,800$ terms.$endgroup$
– Blue
Jan 15 at 13:04