Mixing
The number of prime pairs of $x^2-2y^2=1$
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How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?
I am not getting any clue here.
number-theory prime-numbers contest-math quadratics integers
asked Jan 23 at 19:52
GimgimGimgim
29813
$endgroup$
add a comment |
$begingroup$
How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?
I am not getting any clue here.
number-theory prime-numbers contest-math quadratics integers
asked Jan 23 at 19:52
GimgimGimgim
29813
$endgroup$
1
$begingroup$
Hint: How many odd numbers, when you take them on square and drop one, are not divided by 4?
$endgroup$
– Shaq
Jan 23 at 19:56
1
$begingroup$
This is Pell's Equation
$endgroup$
– lulu
Jan 23 at 19:58
1
$begingroup$
Note: my error. I skipped the requirement that both be prime. In that case, looking at parity suffices.
$endgroup$
– lulu
Jan 23 at 20:02
add a comment |
$begingroup$
How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?
I am not getting any clue here.
number-theory prime-numbers contest-math quadratics integers
asked Jan 23 at 19:52
GimgimGimgim
29813
$endgroup$
How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?
I am not getting any clue here.
number-theory prime-numbers contest-math quadratics integers
number-theory prime-numbers contest-math quadratics integers
asked Jan 23 at 19:52
GimgimGimgim
29813
asked Jan 23 at 19:52
GimgimGimgim
29813
asked Jan 23 at 19:52
GimgimGimgim
29813
asked Jan 23 at 19:52
GimgimGimgim
29813
asked Jan 23 at 19:52
GimgimGimgim
29813
29813
1
$begingroup$
Hint: How many odd numbers, when you take them on square and drop one, are not divided by 4?
$endgroup$
– Shaq
Jan 23 at 19:56
1
$begingroup$
This is Pell's Equation
$endgroup$
– lulu
Jan 23 at 19:58
1
$begingroup$
Note: my error. I skipped the requirement that both be prime. In that case, looking at parity suffices.
$endgroup$
– lulu
Jan 23 at 20:02
add a comment |
1
$begingroup$
Hint: How many odd numbers, when you take them on square and drop one, are not divided by 4?
$endgroup$
– Shaq
Jan 23 at 19:56
1
$begingroup$
This is Pell's Equation
$endgroup$
– lulu
Jan 23 at 19:58
1
$begingroup$
Note: my error. I skipped the requirement that both be prime. In that case, looking at parity suffices.
$endgroup$
– lulu
Jan 23 at 20:02
1
1
$begingroup$
Hint: How many odd numbers, when you take them on square and drop one, are not divided by 4?
$endgroup$
– Shaq
Jan 23 at 19:56
$begingroup$
Hint: How many odd numbers, when you take them on square and drop one, are not divided by 4?
$endgroup$
– Shaq
Jan 23 at 19:56
1
1
$begingroup$
This is Pell's Equation
$endgroup$
– lulu
Jan 23 at 19:58
$begingroup$
This is Pell's Equation
$endgroup$
– lulu
Jan 23 at 19:58
1
1
$begingroup$
Note: my error. I skipped the requirement that both be prime. In that case, looking at parity suffices.
$endgroup$
– lulu
Jan 23 at 20:02
$begingroup$
Note: my error. I skipped the requirement that both be prime. In that case, looking at parity suffices.
$endgroup$
– lulu
Jan 23 at 20:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Flip it around:
$$(x-1)(x+1)=2y^2$$
Because of factor $2$, left hand side must be divisible by $2$. But if one of the factors is divisible by $2$, both are, so overall it's divisible by $4$. Draw your own conclusions from here.
answered Jan 23 at 19:56
orionorion
13.7k11837
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add a comment |
$begingroup$
Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 equiv 1 pmod 4$. If $y$ is odd as well, then $y^2 equiv 1 pmod 4$, too, but $2y^2 equiv 2 pmod 4$, which means that $x^2 - 2y^2 equiv 3 pmod 4$, but clearly $1 equiv 1$, not $3 pmod 4$.
Therefore $y$ must be even, so that $2y^2 equiv 0 pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 pmod 4$ as is necessary to solve the equation.
Of course 2 is prime, and so is $-2$. If $y = pm 2$, then $x = pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Flip it around:
$$(x-1)(x+1)=2y^2$$
Because of factor $2$, left hand side must be divisible by $2$. But if one of the factors is divisible by $2$, both are, so overall it's divisible by $4$. Draw your own conclusions from here.
answered Jan 23 at 19:56
orionorion
13.7k11837
$endgroup$
add a comment |
$begingroup$
Flip it around:
$$(x-1)(x+1)=2y^2$$
Because of factor $2$, left hand side must be divisible by $2$. But if one of the factors is divisible by $2$, both are, so overall it's divisible by $4$. Draw your own conclusions from here.
answered Jan 23 at 19:56
orionorion
13.7k11837
$endgroup$
add a comment |
$begingroup$
Flip it around:
$$(x-1)(x+1)=2y^2$$
Because of factor $2$, left hand side must be divisible by $2$. But if one of the factors is divisible by $2$, both are, so overall it's divisible by $4$. Draw your own conclusions from here.
answered Jan 23 at 19:56
orionorion
13.7k11837
$endgroup$
Flip it around:
$$(x-1)(x+1)=2y^2$$
Because of factor $2$, left hand side must be divisible by $2$. But if one of the factors is divisible by $2$, both are, so overall it's divisible by $4$. Draw your own conclusions from here.
answered Jan 23 at 19:56
orionorion
13.7k11837
answered Jan 23 at 19:56
orionorion
13.7k11837
answered Jan 23 at 19:56
orionorion
13.7k11837
answered Jan 23 at 19:56
orionorion
13.7k11837
13.7k11837
add a comment |
add a comment |
$begingroup$
Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 equiv 1 pmod 4$. If $y$ is odd as well, then $y^2 equiv 1 pmod 4$, too, but $2y^2 equiv 2 pmod 4$, which means that $x^2 - 2y^2 equiv 3 pmod 4$, but clearly $1 equiv 1$, not $3 pmod 4$.
Therefore $y$ must be even, so that $2y^2 equiv 0 pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 pmod 4$ as is necessary to solve the equation.
Of course 2 is prime, and so is $-2$. If $y = pm 2$, then $x = pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
$endgroup$
add a comment |
$begingroup$
Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 equiv 1 pmod 4$. If $y$ is odd as well, then $y^2 equiv 1 pmod 4$, too, but $2y^2 equiv 2 pmod 4$, which means that $x^2 - 2y^2 equiv 3 pmod 4$, but clearly $1 equiv 1$, not $3 pmod 4$.
Therefore $y$ must be even, so that $2y^2 equiv 0 pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 pmod 4$ as is necessary to solve the equation.
Of course 2 is prime, and so is $-2$. If $y = pm 2$, then $x = pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
$endgroup$
add a comment |
$begingroup$
Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 equiv 1 pmod 4$. If $y$ is odd as well, then $y^2 equiv 1 pmod 4$, too, but $2y^2 equiv 2 pmod 4$, which means that $x^2 - 2y^2 equiv 3 pmod 4$, but clearly $1 equiv 1$, not $3 pmod 4$.
Therefore $y$ must be even, so that $2y^2 equiv 0 pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 pmod 4$ as is necessary to solve the equation.
Of course 2 is prime, and so is $-2$. If $y = pm 2$, then $x = pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
$endgroup$
Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 equiv 1 pmod 4$. If $y$ is odd as well, then $y^2 equiv 1 pmod 4$, too, but $2y^2 equiv 2 pmod 4$, which means that $x^2 - 2y^2 equiv 3 pmod 4$, but clearly $1 equiv 1$, not $3 pmod 4$.
Therefore $y$ must be even, so that $2y^2 equiv 0 pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 pmod 4$ as is necessary to solve the equation.
Of course 2 is prime, and so is $-2$. If $y = pm 2$, then $x = pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
answered Jan 25 at 6:42
Robert SoupeRobert Soupe
11.4k21950
11.4k21950
add a comment |
add a comment |
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1
$begingroup$
Hint: How many odd numbers, when you take them on square and drop one, are not divided by 4?
$endgroup$
– Shaq
Jan 23 at 19:56
1
$begingroup$
This is Pell's Equation
$endgroup$
– lulu
Jan 23 at 19:58
1
$begingroup$
Note: my error. I skipped the requirement that both be prime. In that case, looking at parity suffices.
$endgroup$
– lulu
Jan 23 at 20:02