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Next day: apparently my original question is harder, by far, than the other bits. So: it is a finite check, I was able to confirm by computer that, if the polynomial below satisfies $$ f(a,b,c,d) \equiv 0 \pmod {27}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 3, $$ and if $$ f(a,b,c,d) \equiv 0 \pmod {125}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 5, $$

ORIGINAL: $f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at http://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$

Next day: apparently my original question is harder, by far, than the other bits. So: it is a finite check, I was able to confirm by computer that, if the polynomial below satisfies $$ f(a,b,c,d) \equiv 0 \pmod {27}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 3, $$ and if $$ f(a,b,c,d) \equiv 0 \pmod {125}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 5, $$

ORIGINAL: $f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at https://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$

$f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at http://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$

Next day: apparently my original question is harder, by far, than the other bits. So: it is a finite check, I was able to confirm by computer that, if the polynomial below satisfies $$ f(a,b,c,d) \equiv 0 \pmod {27}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 3, $$ and if $$ f(a,b,c,d) \equiv 0 \pmod {125}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 5, $$

ORIGINAL: $f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at http://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$

$f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at http://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EEEEDDDDDDDDDDDIIIIIIIIIIIIIIITTTTTT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EEEEEEEDDDDDDDDDIIIIIIIITTTTTTTT TOOOOOOO: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$

$f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at http://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$