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Primitive Pythagorean Triples. Either a or b must be a multiple of 3
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I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.
We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.
(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = frac(s^2 + t^2)2$ and $b = frac(s^2 - t^2)2$
(5) $a = sqrt(c-b)(c+b) = st$
(6) $a = st$, $b = frac(s^2 - t^2)2$, $c = frac(s^2 + t^2)2$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X neq 0$
(1) $aequiv 0pmod 3$ and $bequiv Xpmod 3$
(2) $bequiv 0pmod 3$ and $aequiv Xpmod 3$
elementary-number-theory pythagorean-triples
edited 2 hours ago
José Carlos Santos
130k17105191
asked 2 hours ago
zono
1235
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up vote
1
down vote
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I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.
We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.
(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = frac(s^2 + t^2)2$ and $b = frac(s^2 - t^2)2$
(5) $a = sqrt(c-b)(c+b) = st$
(6) $a = st$, $b = frac(s^2 - t^2)2$, $c = frac(s^2 + t^2)2$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X neq 0$
(1) $aequiv 0pmod 3$ and $bequiv Xpmod 3$
(2) $bequiv 0pmod 3$ and $aequiv Xpmod 3$
elementary-number-theory pythagorean-triples
edited 2 hours ago
José Carlos Santos
130k17105191
asked 2 hours ago
zono
1235
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
cut-the-knot.org/pythagoras/pythTripleDiv.shtml
– lab bhattacharjee
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.
We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.
(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = frac(s^2 + t^2)2$ and $b = frac(s^2 - t^2)2$
(5) $a = sqrt(c-b)(c+b) = st$
(6) $a = st$, $b = frac(s^2 - t^2)2$, $c = frac(s^2 + t^2)2$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X neq 0$
(1) $aequiv 0pmod 3$ and $bequiv Xpmod 3$
(2) $bequiv 0pmod 3$ and $aequiv Xpmod 3$
elementary-number-theory pythagorean-triples
edited 2 hours ago
José Carlos Santos
130k17105191
asked 2 hours ago
zono
1235
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.
We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.
(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = frac(s^2 + t^2)2$ and $b = frac(s^2 - t^2)2$
(5) $a = sqrt(c-b)(c+b) = st$
(6) $a = st$, $b = frac(s^2 - t^2)2$, $c = frac(s^2 + t^2)2$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X neq 0$
(1) $aequiv 0pmod 3$ and $bequiv Xpmod 3$
(2) $bequiv 0pmod 3$ and $aequiv Xpmod 3$
elementary-number-theory pythagorean-triples
elementary-number-theory pythagorean-triples
edited 2 hours ago
José Carlos Santos
130k17105191
asked 2 hours ago
zono
1235
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
José Carlos Santos
130k17105191
asked 2 hours ago
zono
1235
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
José Carlos Santos
130k17105191
edited 2 hours ago
José Carlos Santos
130k17105191
edited 2 hours ago
José Carlos Santos
130k17105191
130k17105191
asked 2 hours ago
zono
1235
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
zono
1235
asked 2 hours ago
zono
1235
1235
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
zono is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
cut-the-knot.org/pythagoras/pythTripleDiv.shtml
– lab bhattacharjee
1 hour ago
add a comment |Â
1
cut-the-knot.org/pythagoras/pythTripleDiv.shtml
– lab bhattacharjee
1 hour ago
1
1
cut-the-knot.org/pythagoras/pythTripleDiv.shtml
– lab bhattacharjee
1 hour ago
cut-the-knot.org/pythagoras/pythTripleDiv.shtml
– lab bhattacharjee
1 hour ago
add a comment |Â
2 Answers
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oldest
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up vote
5
down vote
The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3kpm1$, for some integer $k$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. Show that this is impossible.
answered 2 hours ago
José Carlos Santos
130k17105191
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
add a comment |Â
up vote
1
down vote
Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
answered 1 hour ago
lab bhattacharjee
217k14153267
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3kpm1$, for some integer $k$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. Show that this is impossible.
answered 2 hours ago
José Carlos Santos
130k17105191
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
add a comment |Â
up vote
5
down vote
The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3kpm1$, for some integer $k$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. Show that this is impossible.
answered 2 hours ago
José Carlos Santos
130k17105191
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3kpm1$, for some integer $k$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. Show that this is impossible.
answered 2 hours ago
José Carlos Santos
130k17105191
The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3kpm1$, for some integer $k$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. Show that this is impossible.
answered 2 hours ago
José Carlos Santos
130k17105191
answered 2 hours ago
José Carlos Santos
130k17105191
answered 2 hours ago
José Carlos Santos
130k17105191
answered 2 hours ago
José Carlos Santos
130k17105191
130k17105191
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
add a comment |Â
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
ooh! I should have used proof by contradiction. I understand the first one (both be multiples of 3) . Now still I'm still thinking about the second one why the form is 3k±1..
– zono
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
If $a$ is not a multiple of $3$ then, since the $3$ numbers $a-1$, $a$, and $a+1$ are consecutive and therefore one of them is a multiple of $3$, then $a-1$ or $a+1$ is a multiple of $3$.
– José Carlos Santos
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
Thank you! I understand it. If possible.. can you tell me a little bit more about "3k+2". Although I took 15 min to think about it but I could not get it.. "c^2(=a^2+b^2) is of the form 3k+2"
– zono
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
You don't understand why is it that when I add two numers which are a multiple of $3$ plus $1$, then what I get is a multiple of $3$ plus $2$?
– José Carlos Santos
1 hour ago
add a comment |Â
up vote
1
down vote
Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
answered 1 hour ago
lab bhattacharjee
217k14153267
add a comment |Â
up vote
1
down vote
Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
answered 1 hour ago
lab bhattacharjee
217k14153267
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
answered 1 hour ago
lab bhattacharjee
217k14153267
Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
answered 1 hour ago
lab bhattacharjee
217k14153267
answered 1 hour ago
lab bhattacharjee
217k14153267
answered 1 hour ago
lab bhattacharjee
217k14153267
answered 1 hour ago
lab bhattacharjee
217k14153267
217k14153267
add a comment |Â
add a comment |Â
zono is a new contributor. Be nice, and check out our Code of Conduct.
zono is a new contributor. Be nice, and check out our Code of Conduct.
zono is a new contributor. Be nice, and check out our Code of Conduct.
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1
cut-the-knot.org/pythagoras/pythTripleDiv.shtml
– lab bhattacharjee
1 hour ago