Number of Pythagorean Triples
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I am trying to solve an exercise from the book "Theory of Numbers" by B.M.Stewart. The exercise is the following one:
Let $T=2^ap_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, where $a ge0, nge0, 2<p_1<p_2<dots p_n, p_j$ odd prime numbers $ forall j=1 dots n,a_j ge1$ and let $S(T)$ indicate the number of primitive Pythagorean triplets of side $T$. Show that $S(T) = 2^{n-1}$ if $a=0$.
The primitive Pythagorean triplets are the solutions of $x^2+y^2=z^2$ where $gcd(x,y,z)=1$ and they are given by $$cases {x=2uv \y=u^2-v^2\z=u^2+v^2\u>v>0\ gcd(u,v)=1\ u notequiv v pmod2} $$
If $a=0$ then $T=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, so $T ne 2uv$. Then $T=y$ or $T=z$.
If $T=y$, then $$T=u^2-v^2=(u+v)(u-v)=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}.$$ So for $(u+v)$ I have $2^n$ possibilities because I can count the number of prime factors which is $tau(r)$ with $r=p_1p_2 dots p_n$.
From there I don't know how to proceed. Have you any idea?
number-theory pythagorean-triples divisor-counting-function
asked Jan 24 at 15:55
Phi_24Phi_24
1938
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add a comment |
$begingroup$
I am trying to solve an exercise from the book "Theory of Numbers" by B.M.Stewart. The exercise is the following one:
Let $T=2^ap_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, where $a ge0, nge0, 2<p_1<p_2<dots p_n, p_j$ odd prime numbers $ forall j=1 dots n,a_j ge1$ and let $S(T)$ indicate the number of primitive Pythagorean triplets of side $T$. Show that $S(T) = 2^{n-1}$ if $a=0$.
The primitive Pythagorean triplets are the solutions of $x^2+y^2=z^2$ where $gcd(x,y,z)=1$ and they are given by $$cases {x=2uv \y=u^2-v^2\z=u^2+v^2\u>v>0\ gcd(u,v)=1\ u notequiv v pmod2} $$
If $a=0$ then $T=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, so $T ne 2uv$. Then $T=y$ or $T=z$.
If $T=y$, then $$T=u^2-v^2=(u+v)(u-v)=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}.$$ So for $(u+v)$ I have $2^n$ possibilities because I can count the number of prime factors which is $tau(r)$ with $r=p_1p_2 dots p_n$.
From there I don't know how to proceed. Have you any idea?
number-theory pythagorean-triples divisor-counting-function
asked Jan 24 at 15:55
Phi_24Phi_24
1938
$endgroup$
$begingroup$
I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$
$endgroup$
– saulspatz
Jan 24 at 16:26
add a comment |
$begingroup$
I am trying to solve an exercise from the book "Theory of Numbers" by B.M.Stewart. The exercise is the following one:
Let $T=2^ap_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, where $a ge0, nge0, 2<p_1<p_2<dots p_n, p_j$ odd prime numbers $ forall j=1 dots n,a_j ge1$ and let $S(T)$ indicate the number of primitive Pythagorean triplets of side $T$. Show that $S(T) = 2^{n-1}$ if $a=0$.
The primitive Pythagorean triplets are the solutions of $x^2+y^2=z^2$ where $gcd(x,y,z)=1$ and they are given by $$cases {x=2uv \y=u^2-v^2\z=u^2+v^2\u>v>0\ gcd(u,v)=1\ u notequiv v pmod2} $$
If $a=0$ then $T=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, so $T ne 2uv$. Then $T=y$ or $T=z$.
If $T=y$, then $$T=u^2-v^2=(u+v)(u-v)=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}.$$ So for $(u+v)$ I have $2^n$ possibilities because I can count the number of prime factors which is $tau(r)$ with $r=p_1p_2 dots p_n$.
From there I don't know how to proceed. Have you any idea?
number-theory pythagorean-triples divisor-counting-function
asked Jan 24 at 15:55
Phi_24Phi_24
1938
$endgroup$
I am trying to solve an exercise from the book "Theory of Numbers" by B.M.Stewart. The exercise is the following one:
Let $T=2^ap_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, where $a ge0, nge0, 2<p_1<p_2<dots p_n, p_j$ odd prime numbers $ forall j=1 dots n,a_j ge1$ and let $S(T)$ indicate the number of primitive Pythagorean triplets of side $T$. Show that $S(T) = 2^{n-1}$ if $a=0$.
The primitive Pythagorean triplets are the solutions of $x^2+y^2=z^2$ where $gcd(x,y,z)=1$ and they are given by $$cases {x=2uv \y=u^2-v^2\z=u^2+v^2\u>v>0\ gcd(u,v)=1\ u notequiv v pmod2} $$
If $a=0$ then $T=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}$, so $T ne 2uv$. Then $T=y$ or $T=z$.
If $T=y$, then $$T=u^2-v^2=(u+v)(u-v)=p_1^{a_1}p_2^{a_2} dots p_n^{a_n}.$$ So for $(u+v)$ I have $2^n$ possibilities because I can count the number of prime factors which is $tau(r)$ with $r=p_1p_2 dots p_n$.
From there I don't know how to proceed. Have you any idea?
number-theory pythagorean-triples divisor-counting-function
number-theory pythagorean-triples divisor-counting-function
asked Jan 24 at 15:55
Phi_24Phi_24
1938
asked Jan 24 at 15:55
Phi_24Phi_24
1938
asked Jan 24 at 15:55
Phi_24Phi_24
1938
asked Jan 24 at 15:55
Phi_24Phi_24
1938
asked Jan 24 at 15:55
Phi_24Phi_24
1938
1938
$begingroup$
I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$
$endgroup$
– saulspatz
Jan 24 at 16:26
add a comment |
$begingroup$
I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$
$endgroup$
– saulspatz
Jan 24 at 16:26
$begingroup$
I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$
$endgroup$
– saulspatz
Jan 24 at 16:26
$begingroup$
I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$
$endgroup$
– saulspatz
Jan 24 at 16:26
add a comment |
1 Answer
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You're awfully close, if you accept my comment that only the $T=y$ case needs to be considered. It follows from $gcd(u,v)=1, u notequiv v pmod2$ that $gcd(u+v,u-v)=1.$ There are $2^n$ ways to split $T$ up into a pair of relative prime factors, and the larger one must be $u+v,$ so you just solve two linear equations for $u$ and $v$.
EDIT Suppose we had $T=3^2cdot5cdot7.$ We could split this into two the factors $9$ and $35$. Then we would have to solve $$begin{align}u+v &=35\u-v&=9end{align}$$ so $u=22,v=13.$
We have to divide the number of solutions by $2$ because we could always have to make $u+v$ the larger factor, so half the choices are inadmissible.
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
$endgroup$
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
add a comment |
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1 Answer
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You're awfully close, if you accept my comment that only the $T=y$ case needs to be considered. It follows from $gcd(u,v)=1, u notequiv v pmod2$ that $gcd(u+v,u-v)=1.$ There are $2^n$ ways to split $T$ up into a pair of relative prime factors, and the larger one must be $u+v,$ so you just solve two linear equations for $u$ and $v$.
EDIT Suppose we had $T=3^2cdot5cdot7.$ We could split this into two the factors $9$ and $35$. Then we would have to solve $$begin{align}u+v &=35\u-v&=9end{align}$$ so $u=22,v=13.$
We have to divide the number of solutions by $2$ because we could always have to make $u+v$ the larger factor, so half the choices are inadmissible.
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
$endgroup$
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
add a comment |
$begingroup$
You're awfully close, if you accept my comment that only the $T=y$ case needs to be considered. It follows from $gcd(u,v)=1, u notequiv v pmod2$ that $gcd(u+v,u-v)=1.$ There are $2^n$ ways to split $T$ up into a pair of relative prime factors, and the larger one must be $u+v,$ so you just solve two linear equations for $u$ and $v$.
EDIT Suppose we had $T=3^2cdot5cdot7.$ We could split this into two the factors $9$ and $35$. Then we would have to solve $$begin{align}u+v &=35\u-v&=9end{align}$$ so $u=22,v=13.$
We have to divide the number of solutions by $2$ because we could always have to make $u+v$ the larger factor, so half the choices are inadmissible.
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
$endgroup$
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
add a comment |
$begingroup$
You're awfully close, if you accept my comment that only the $T=y$ case needs to be considered. It follows from $gcd(u,v)=1, u notequiv v pmod2$ that $gcd(u+v,u-v)=1.$ There are $2^n$ ways to split $T$ up into a pair of relative prime factors, and the larger one must be $u+v,$ so you just solve two linear equations for $u$ and $v$.
EDIT Suppose we had $T=3^2cdot5cdot7.$ We could split this into two the factors $9$ and $35$. Then we would have to solve $$begin{align}u+v &=35\u-v&=9end{align}$$ so $u=22,v=13.$
We have to divide the number of solutions by $2$ because we could always have to make $u+v$ the larger factor, so half the choices are inadmissible.
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
$endgroup$
You're awfully close, if you accept my comment that only the $T=y$ case needs to be considered. It follows from $gcd(u,v)=1, u notequiv v pmod2$ that $gcd(u+v,u-v)=1.$ There are $2^n$ ways to split $T$ up into a pair of relative prime factors, and the larger one must be $u+v,$ so you just solve two linear equations for $u$ and $v$.
EDIT Suppose we had $T=3^2cdot5cdot7.$ We could split this into two the factors $9$ and $35$. Then we would have to solve $$begin{align}u+v &=35\u-v&=9end{align}$$ so $u=22,v=13.$
We have to divide the number of solutions by $2$ because we could always have to make $u+v$ the larger factor, so half the choices are inadmissible.
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
edited Jan 24 at 17:54
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
answered Jan 24 at 16:39
saulspatzsaulspatz
16.6k31333
16.6k31333
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
add a comment |
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
$begingroup$
I'm sorry but I think I am not able to see the linear equations. Is the solution $2^{n-1}$ because $u-v neq 1$ (I don't see why) and so I can only make $2^{n-1}$ choices for $u+v$?
$endgroup$
– Phi_24
Jan 24 at 17:38
add a comment |
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$begingroup$
I think by "side" they mean "leg," that is, not the hypotenuse. Otherwise, consider $T=5.$ The formula gives one triple, but we have $3,4,5$ and $5,12,13.$ So I think you don't need to consider $T=z.$
$endgroup$
– saulspatz
Jan 24 at 16:26