Mixing
Class Number Calculation of a Real Quadratic Number Field
$begingroup$
I am looking at the example below. Can anyone explain how they end up with the contradiction. Why do they reduce $a^2-65b^2$ modulo $5$ to show that it has no integer solutions?
number-theory algebraic-number-theory ideals maximal-and-prime-ideals
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
$endgroup$
add a comment |
$begingroup$
I am looking at the example below. Can anyone explain how they end up with the contradiction. Why do they reduce $a^2-65b^2$ modulo $5$ to show that it has no integer solutions?
number-theory algebraic-number-theory ideals maximal-and-prime-ideals
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
$endgroup$
4
$begingroup$
The congruences $x^2equivpm8pmod5$ are insoluble.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 17:38
add a comment |
$begingroup$
I am looking at the example below. Can anyone explain how they end up with the contradiction. Why do they reduce $a^2-65b^2$ modulo $5$ to show that it has no integer solutions?
number-theory algebraic-number-theory ideals maximal-and-prime-ideals
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
$endgroup$
I am looking at the example below. Can anyone explain how they end up with the contradiction. Why do they reduce $a^2-65b^2$ modulo $5$ to show that it has no integer solutions?
number-theory algebraic-number-theory ideals maximal-and-prime-ideals
number-theory algebraic-number-theory ideals maximal-and-prime-ideals
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
asked Jan 10 at 16:52
pullofthemoonpullofthemoon
815
815
4
$begingroup$
The congruences $x^2equivpm8pmod5$ are insoluble.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 17:38
add a comment |
4
$begingroup$
The congruences $x^2equivpm8pmod5$ are insoluble.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 17:38
4
4
$begingroup$
The congruences $x^2equivpm8pmod5$ are insoluble.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 17:38
$begingroup$
The congruences $x^2equivpm8pmod5$ are insoluble.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 17:38
add a comment |
2 Answers
2
active
oldest
votes
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It's mostly a matter of efficiency. Try it modulo 65. Obviously $65b^2 equiv 0 pmod{65}$. So we need to look just at $a^2 pmod{65}$. But we need to look at 65 values: 0, 1, 4, 9, 16, 25, 36, 49, 64, 16, 35, 56, 14, 39, 1, 30, 61, 29, 64, 36, 10, 51, 29, 9, 56, 40, 26, 14, 4, 61, 55, 51, 49, 49, 51, 55, 61, 4, 14, 26, 40, 56, 9, 29, 51, 10, 36, 64, 29, 61, 30, 1, 39, 14, 56, 35, 16, 64, 49, 36, 25, 16, 9, 4, 1.
If instead we look at it modulo 5, it still zeroes out $65b^2$ but we only have five values of $a^2$ to worry about, namely 0, 1, 4, 4, 1. It is much easier to see that $pm 2$, $pm 8$ don't occur here.
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
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$begingroup$
Reducing $a^2-65b^2=pm 8$ modulo $5$ gives $$a^2equiv pm 2bmod 5,$$ so $left( frac{2}{5}right)=left( frac{3}{5}right)=1$ for the Legendre symbols. On the other hand, there is no square $a^2$ equal to $pm 2$ in $Bbb F_5$, just test: $0^2=0$, $1^2=1$, $2^2=4$, $3^2=4$ and $4^2=1$. This is a contradiction.
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
1
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
It's mostly a matter of efficiency. Try it modulo 65. Obviously $65b^2 equiv 0 pmod{65}$. So we need to look just at $a^2 pmod{65}$. But we need to look at 65 values: 0, 1, 4, 9, 16, 25, 36, 49, 64, 16, 35, 56, 14, 39, 1, 30, 61, 29, 64, 36, 10, 51, 29, 9, 56, 40, 26, 14, 4, 61, 55, 51, 49, 49, 51, 55, 61, 4, 14, 26, 40, 56, 9, 29, 51, 10, 36, 64, 29, 61, 30, 1, 39, 14, 56, 35, 16, 64, 49, 36, 25, 16, 9, 4, 1.
If instead we look at it modulo 5, it still zeroes out $65b^2$ but we only have five values of $a^2$ to worry about, namely 0, 1, 4, 4, 1. It is much easier to see that $pm 2$, $pm 8$ don't occur here.
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
$endgroup$
add a comment |
$begingroup$
It's mostly a matter of efficiency. Try it modulo 65. Obviously $65b^2 equiv 0 pmod{65}$. So we need to look just at $a^2 pmod{65}$. But we need to look at 65 values: 0, 1, 4, 9, 16, 25, 36, 49, 64, 16, 35, 56, 14, 39, 1, 30, 61, 29, 64, 36, 10, 51, 29, 9, 56, 40, 26, 14, 4, 61, 55, 51, 49, 49, 51, 55, 61, 4, 14, 26, 40, 56, 9, 29, 51, 10, 36, 64, 29, 61, 30, 1, 39, 14, 56, 35, 16, 64, 49, 36, 25, 16, 9, 4, 1.
If instead we look at it modulo 5, it still zeroes out $65b^2$ but we only have five values of $a^2$ to worry about, namely 0, 1, 4, 4, 1. It is much easier to see that $pm 2$, $pm 8$ don't occur here.
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
$endgroup$
add a comment |
$begingroup$
It's mostly a matter of efficiency. Try it modulo 65. Obviously $65b^2 equiv 0 pmod{65}$. So we need to look just at $a^2 pmod{65}$. But we need to look at 65 values: 0, 1, 4, 9, 16, 25, 36, 49, 64, 16, 35, 56, 14, 39, 1, 30, 61, 29, 64, 36, 10, 51, 29, 9, 56, 40, 26, 14, 4, 61, 55, 51, 49, 49, 51, 55, 61, 4, 14, 26, 40, 56, 9, 29, 51, 10, 36, 64, 29, 61, 30, 1, 39, 14, 56, 35, 16, 64, 49, 36, 25, 16, 9, 4, 1.
If instead we look at it modulo 5, it still zeroes out $65b^2$ but we only have five values of $a^2$ to worry about, namely 0, 1, 4, 4, 1. It is much easier to see that $pm 2$, $pm 8$ don't occur here.
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
$endgroup$
It's mostly a matter of efficiency. Try it modulo 65. Obviously $65b^2 equiv 0 pmod{65}$. So we need to look just at $a^2 pmod{65}$. But we need to look at 65 values: 0, 1, 4, 9, 16, 25, 36, 49, 64, 16, 35, 56, 14, 39, 1, 30, 61, 29, 64, 36, 10, 51, 29, 9, 56, 40, 26, 14, 4, 61, 55, 51, 49, 49, 51, 55, 61, 4, 14, 26, 40, 56, 9, 29, 51, 10, 36, 64, 29, 61, 30, 1, 39, 14, 56, 35, 16, 64, 49, 36, 25, 16, 9, 4, 1.
If instead we look at it modulo 5, it still zeroes out $65b^2$ but we only have five values of $a^2$ to worry about, namely 0, 1, 4, 4, 1. It is much easier to see that $pm 2$, $pm 8$ don't occur here.
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
answered Jan 13 at 1:36
Robert SoupeRobert Soupe
11.1k21950
11.1k21950
add a comment |
add a comment |
$begingroup$
Reducing $a^2-65b^2=pm 8$ modulo $5$ gives $$a^2equiv pm 2bmod 5,$$ so $left( frac{2}{5}right)=left( frac{3}{5}right)=1$ for the Legendre symbols. On the other hand, there is no square $a^2$ equal to $pm 2$ in $Bbb F_5$, just test: $0^2=0$, $1^2=1$, $2^2=4$, $3^2=4$ and $4^2=1$. This is a contradiction.
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
1
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
add a comment |
$begingroup$
Reducing $a^2-65b^2=pm 8$ modulo $5$ gives $$a^2equiv pm 2bmod 5,$$ so $left( frac{2}{5}right)=left( frac{3}{5}right)=1$ for the Legendre symbols. On the other hand, there is no square $a^2$ equal to $pm 2$ in $Bbb F_5$, just test: $0^2=0$, $1^2=1$, $2^2=4$, $3^2=4$ and $4^2=1$. This is a contradiction.
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
1
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
add a comment |
$begingroup$
Reducing $a^2-65b^2=pm 8$ modulo $5$ gives $$a^2equiv pm 2bmod 5,$$ so $left( frac{2}{5}right)=left( frac{3}{5}right)=1$ for the Legendre symbols. On the other hand, there is no square $a^2$ equal to $pm 2$ in $Bbb F_5$, just test: $0^2=0$, $1^2=1$, $2^2=4$, $3^2=4$ and $4^2=1$. This is a contradiction.
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
Reducing $a^2-65b^2=pm 8$ modulo $5$ gives $$a^2equiv pm 2bmod 5,$$ so $left( frac{2}{5}right)=left( frac{3}{5}right)=1$ for the Legendre symbols. On the other hand, there is no square $a^2$ equal to $pm 2$ in $Bbb F_5$, just test: $0^2=0$, $1^2=1$, $2^2=4$, $3^2=4$ and $4^2=1$. This is a contradiction.
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
answered Jan 10 at 20:09
Dietrich BurdeDietrich Burde
79k647102
79k647102
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
1
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
add a comment |
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
1
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
$begingroup$
That’s great! Thank you! Can I just ask why are reducing mod 5 to find integer solutions? What is the reason for this?
$endgroup$
– pullofthemoon
Jan 10 at 22:29
1
1
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
$begingroup$
Because reducing modulo $5$ removes the $65$ from the picture, that’s all.
$endgroup$
– Lubin
Jan 10 at 23:07
add a comment |
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$begingroup$
The congruences $x^2equivpm8pmod5$ are insoluble.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 17:38