On largest box expected size not containing integer vector solutions
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I am trying to understand the largest cube in $Bbb Z^8$ around origin not containing integer vector $(a,b,c,d,a',b',c',d')$ solutions to $$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$ where $A,B,C,D,E,F$ are pairwise coprime where $$n<A,B,C,D,E,F<2n$$ $$n/8<|A-B|,|C-D|,|A-C|,|A-D|,|B-C|,|B-D|,|A-E|,|B-E|,dots,|E-F|<n/4$$ holds?
Is the box $alphacdot n^{3/7}$ in length around origin for some constant $0<alpha $ with probability $1-o(1)$? That is a vector $(a,b,c,d,a',b',c',d')inBbb Z^8$ with $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ exists but with probability $1-o(1)$ no much smaller one exists?
This is where I get the $3/7$ value.
$$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$
$$implies E(aBD+bBC+cAD+dAC)=-F(a'BD+b'BC+c'AD+d'AC)$$
which gives
$$a'BD+b'BC+c'AD+d'AC=ell E$$
$$aBD+bBC+cAD+dAC=-ell F$$
If size of $ell$ is $n^eta$ then size of $ell E$ is size of $a'BD+b'BC+c'AD+d'AC$ is $n^{2+gamma}$. We have $n^{4gamma}$ values in $a'b'c'd'$ and $n^{4gamma}$ values in $abcd$ and we want to hit $ell E$ and $ell F$ (a total of $2n^{eta}$ possibilities) of a total of $n^{2(2+gamma)}$ possibilities. The expected intersection is
$$frac{n^{4gamma}n^{4gamma}2n^{eta}}{n^{2(2+gamma)}}=frac{2n^{6gamma+eta}}{n^4}.$$
We also know roughly $2+gamma=1+eta$ by equating sizes which implies $eta-gamma=1$. Thus intersection is $$frac{2n^{6gamma+eta}}{n^4}asympfrac{2n^{6gamma+gamma+1}}{n^4}$$ which gives $gamma>frac37$ and thus we expect $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ to hold.
linear-algebra probability diophantine-equations integers linear-diophantine-equations
asked Jan 8 at 16:12
BroutBrout
2,5591431
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add a comment |
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I am trying to understand the largest cube in $Bbb Z^8$ around origin not containing integer vector $(a,b,c,d,a',b',c',d')$ solutions to $$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$ where $A,B,C,D,E,F$ are pairwise coprime where $$n<A,B,C,D,E,F<2n$$ $$n/8<|A-B|,|C-D|,|A-C|,|A-D|,|B-C|,|B-D|,|A-E|,|B-E|,dots,|E-F|<n/4$$ holds?
Is the box $alphacdot n^{3/7}$ in length around origin for some constant $0<alpha $ with probability $1-o(1)$? That is a vector $(a,b,c,d,a',b',c',d')inBbb Z^8$ with $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ exists but with probability $1-o(1)$ no much smaller one exists?
This is where I get the $3/7$ value.
$$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$
$$implies E(aBD+bBC+cAD+dAC)=-F(a'BD+b'BC+c'AD+d'AC)$$
which gives
$$a'BD+b'BC+c'AD+d'AC=ell E$$
$$aBD+bBC+cAD+dAC=-ell F$$
If size of $ell$ is $n^eta$ then size of $ell E$ is size of $a'BD+b'BC+c'AD+d'AC$ is $n^{2+gamma}$. We have $n^{4gamma}$ values in $a'b'c'd'$ and $n^{4gamma}$ values in $abcd$ and we want to hit $ell E$ and $ell F$ (a total of $2n^{eta}$ possibilities) of a total of $n^{2(2+gamma)}$ possibilities. The expected intersection is
$$frac{n^{4gamma}n^{4gamma}2n^{eta}}{n^{2(2+gamma)}}=frac{2n^{6gamma+eta}}{n^4}.$$
We also know roughly $2+gamma=1+eta$ by equating sizes which implies $eta-gamma=1$. Thus intersection is $$frac{2n^{6gamma+eta}}{n^4}asympfrac{2n^{6gamma+gamma+1}}{n^4}$$ which gives $gamma>frac37$ and thus we expect $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ to hold.
linear-algebra probability diophantine-equations integers linear-diophantine-equations
asked Jan 8 at 16:12
BroutBrout
2,5591431
$endgroup$
add a comment |
$begingroup$
I am trying to understand the largest cube in $Bbb Z^8$ around origin not containing integer vector $(a,b,c,d,a',b',c',d')$ solutions to $$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$ where $A,B,C,D,E,F$ are pairwise coprime where $$n<A,B,C,D,E,F<2n$$ $$n/8<|A-B|,|C-D|,|A-C|,|A-D|,|B-C|,|B-D|,|A-E|,|B-E|,dots,|E-F|<n/4$$ holds?
Is the box $alphacdot n^{3/7}$ in length around origin for some constant $0<alpha $ with probability $1-o(1)$? That is a vector $(a,b,c,d,a',b',c',d')inBbb Z^8$ with $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ exists but with probability $1-o(1)$ no much smaller one exists?
This is where I get the $3/7$ value.
$$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$
$$implies E(aBD+bBC+cAD+dAC)=-F(a'BD+b'BC+c'AD+d'AC)$$
which gives
$$a'BD+b'BC+c'AD+d'AC=ell E$$
$$aBD+bBC+cAD+dAC=-ell F$$
If size of $ell$ is $n^eta$ then size of $ell E$ is size of $a'BD+b'BC+c'AD+d'AC$ is $n^{2+gamma}$. We have $n^{4gamma}$ values in $a'b'c'd'$ and $n^{4gamma}$ values in $abcd$ and we want to hit $ell E$ and $ell F$ (a total of $2n^{eta}$ possibilities) of a total of $n^{2(2+gamma)}$ possibilities. The expected intersection is
$$frac{n^{4gamma}n^{4gamma}2n^{eta}}{n^{2(2+gamma)}}=frac{2n^{6gamma+eta}}{n^4}.$$
We also know roughly $2+gamma=1+eta$ by equating sizes which implies $eta-gamma=1$. Thus intersection is $$frac{2n^{6gamma+eta}}{n^4}asympfrac{2n^{6gamma+gamma+1}}{n^4}$$ which gives $gamma>frac37$ and thus we expect $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ to hold.
linear-algebra probability diophantine-equations integers linear-diophantine-equations
asked Jan 8 at 16:12
BroutBrout
2,5591431
$endgroup$
I am trying to understand the largest cube in $Bbb Z^8$ around origin not containing integer vector $(a,b,c,d,a',b',c',d')$ solutions to $$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$ where $A,B,C,D,E,F$ are pairwise coprime where $$n<A,B,C,D,E,F<2n$$ $$n/8<|A-B|,|C-D|,|A-C|,|A-D|,|B-C|,|B-D|,|A-E|,|B-E|,dots,|E-F|<n/4$$ holds?
Is the box $alphacdot n^{3/7}$ in length around origin for some constant $0<alpha $ with probability $1-o(1)$? That is a vector $(a,b,c,d,a',b',c',d')inBbb Z^8$ with $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ exists but with probability $1-o(1)$ no much smaller one exists?
This is where I get the $3/7$ value.
$$aBDE+bBCE+cADE+dACE+a'BDF+b'BCF+c'ADF+d'ACF=0$$
$$implies E(aBD+bBC+cAD+dAC)=-F(a'BD+b'BC+c'AD+d'AC)$$
which gives
$$a'BD+b'BC+c'AD+d'AC=ell E$$
$$aBD+bBC+cAD+dAC=-ell F$$
If size of $ell$ is $n^eta$ then size of $ell E$ is size of $a'BD+b'BC+c'AD+d'AC$ is $n^{2+gamma}$. We have $n^{4gamma}$ values in $a'b'c'd'$ and $n^{4gamma}$ values in $abcd$ and we want to hit $ell E$ and $ell F$ (a total of $2n^{eta}$ possibilities) of a total of $n^{2(2+gamma)}$ possibilities. The expected intersection is
$$frac{n^{4gamma}n^{4gamma}2n^{eta}}{n^{2(2+gamma)}}=frac{2n^{6gamma+eta}}{n^4}.$$
We also know roughly $2+gamma=1+eta$ by equating sizes which implies $eta-gamma=1$. Thus intersection is $$frac{2n^{6gamma+eta}}{n^4}asympfrac{2n^{6gamma+gamma+1}}{n^4}$$ which gives $gamma>frac37$ and thus we expect $max(|a|,|b|,|c|,|d|,|a'|,|b'|,|c'|,|d'|)asymp n^{3/7}$ to hold.
linear-algebra probability diophantine-equations integers linear-diophantine-equations
linear-algebra probability diophantine-equations integers linear-diophantine-equations
asked Jan 8 at 16:12
BroutBrout
2,5591431
asked Jan 8 at 16:12
BroutBrout
2,5591431
edited Jan 8 at 16:23
Brout
asked Jan 8 at 16:12
BroutBrout
2,5591431
asked Jan 8 at 16:12
BroutBrout
2,5591431
asked Jan 8 at 16:12
BroutBrout
2,5591431
2,5591431
add a comment |
add a comment |
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