Mixing
Is there integer $x$ such that $79|7x^2+4x-23$
$begingroup$
Is there integer $x$ such that $79|7x^2+4x-23$ ?
I keep getting that there is $x$ that satisfies this condition, but online calculator keeps saying that there is not.
I worked it out using Legendre's symbol:
If $y=7x+2$, then starting equation is equivalent to $y^2 equiv 7$ mod$79$,
and because $genfrac{(}{)}{}{}{7}{79} = genfrac{(}{)}{}{}{79}{7} = 1$, equation has a solution ?
elementary-number-theory quadratic-residues legendre-symbol
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
Is there integer $x$ such that $79|7x^2+4x-23$ ?
I keep getting that there is $x$ that satisfies this condition, but online calculator keeps saying that there is not.
I worked it out using Legendre's symbol:
If $y=7x+2$, then starting equation is equivalent to $y^2 equiv 7$ mod$79$,
and because $genfrac{(}{)}{}{}{7}{79} = genfrac{(}{)}{}{}{79}{7} = 1$, equation has a solution ?
elementary-number-theory quadratic-residues legendre-symbol
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
$endgroup$
$begingroup$
Possible duplicate of Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$
$endgroup$
– Bill Dubuque
Jan 25 at 15:44
add a comment |
$begingroup$
Is there integer $x$ such that $79|7x^2+4x-23$ ?
I keep getting that there is $x$ that satisfies this condition, but online calculator keeps saying that there is not.
I worked it out using Legendre's symbol:
If $y=7x+2$, then starting equation is equivalent to $y^2 equiv 7$ mod$79$,
and because $genfrac{(}{)}{}{}{7}{79} = genfrac{(}{)}{}{}{79}{7} = 1$, equation has a solution ?
elementary-number-theory quadratic-residues legendre-symbol
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
$endgroup$
Is there integer $x$ such that $79|7x^2+4x-23$ ?
I keep getting that there is $x$ that satisfies this condition, but online calculator keeps saying that there is not.
I worked it out using Legendre's symbol:
If $y=7x+2$, then starting equation is equivalent to $y^2 equiv 7$ mod$79$,
and because $genfrac{(}{)}{}{}{7}{79} = genfrac{(}{)}{}{}{79}{7} = 1$, equation has a solution ?
elementary-number-theory quadratic-residues legendre-symbol
elementary-number-theory quadratic-residues legendre-symbol
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
edited Jan 25 at 15:44
Bill Dubuque
212k29195654
212k29195654
asked Jan 25 at 14:57
user626177
$begingroup$
Possible duplicate of Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$
$endgroup$
– Bill Dubuque
Jan 25 at 15:44
add a comment |
$begingroup$
Possible duplicate of Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$
$endgroup$
– Bill Dubuque
Jan 25 at 15:44
$begingroup$
Possible duplicate of Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$
$endgroup$
– Bill Dubuque
Jan 25 at 15:44
$begingroup$
Possible duplicate of Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$
$endgroup$
– Bill Dubuque
Jan 25 at 15:44
add a comment |
1 Answer
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$begingroup$
Note that by quadratic reciprocity
$$
left(frac{7}{79}right)=(-1)^{3cdot39}left(frac{79}{7}right)=-left(frac{2}{7}right)=-1.
$$
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
$endgroup$
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Note that by quadratic reciprocity
$$
left(frac{7}{79}right)=(-1)^{3cdot39}left(frac{79}{7}right)=-left(frac{2}{7}right)=-1.
$$
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
$endgroup$
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
add a comment |
$begingroup$
Note that by quadratic reciprocity
$$
left(frac{7}{79}right)=(-1)^{3cdot39}left(frac{79}{7}right)=-left(frac{2}{7}right)=-1.
$$
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
$endgroup$
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
add a comment |
$begingroup$
Note that by quadratic reciprocity
$$
left(frac{7}{79}right)=(-1)^{3cdot39}left(frac{79}{7}right)=-left(frac{2}{7}right)=-1.
$$
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
$endgroup$
Note that by quadratic reciprocity
$$
left(frac{7}{79}right)=(-1)^{3cdot39}left(frac{79}{7}right)=-left(frac{2}{7}right)=-1.
$$
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
edited Jan 25 at 15:11
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
answered Jan 25 at 15:09
studiosusstudiosus
2,174715
2,174715
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
add a comment |
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
$begingroup$
Oh my god .....
$endgroup$
– user626177
Jan 25 at 15:10
add a comment |
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$begingroup$
Possible duplicate of Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$
$endgroup$
– Bill Dubuque
Jan 25 at 15:44