Odd Pythagorian triplets [closed]
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How many Pythagorean triplets ${a,b,c}$ exist, where $a,b,c$ are all odd?
As far as I know there are no such triplets. ${3, 4, 5}; {5,12,13} ; {7,24,25}$ and its multiples are examples.
Is there any explanation on why all the integers forming a triplet are not all odd or all even?
algebra-precalculus elementary-number-theory euclidean-geometry pythagorean-triples
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
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closed as off-topic by Namaste, Holo, Cesareo, Saad, Did Dec 25 '18 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Cesareo, Saad, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How many Pythagorean triplets ${a,b,c}$ exist, where $a,b,c$ are all odd?
As far as I know there are no such triplets. ${3, 4, 5}; {5,12,13} ; {7,24,25}$ and its multiples are examples.
Is there any explanation on why all the integers forming a triplet are not all odd or all even?
algebra-precalculus elementary-number-theory euclidean-geometry pythagorean-triples
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
$endgroup$
closed as off-topic by Namaste, Holo, Cesareo, Saad, Did Dec 25 '18 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Cesareo, Saad, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
3
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Think about sums and squares of odds and evens.
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– Randall
Dec 22 '18 at 15:24
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Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic
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– Fathima Jasmine
Dec 22 '18 at 15:31
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All even: 6, 8, 10.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:55
add a comment |
$begingroup$
How many Pythagorean triplets ${a,b,c}$ exist, where $a,b,c$ are all odd?
As far as I know there are no such triplets. ${3, 4, 5}; {5,12,13} ; {7,24,25}$ and its multiples are examples.
Is there any explanation on why all the integers forming a triplet are not all odd or all even?
algebra-precalculus elementary-number-theory euclidean-geometry pythagorean-triples
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
$endgroup$
How many Pythagorean triplets ${a,b,c}$ exist, where $a,b,c$ are all odd?
As far as I know there are no such triplets. ${3, 4, 5}; {5,12,13} ; {7,24,25}$ and its multiples are examples.
Is there any explanation on why all the integers forming a triplet are not all odd or all even?
algebra-precalculus elementary-number-theory euclidean-geometry pythagorean-triples
algebra-precalculus elementary-number-theory euclidean-geometry pythagorean-triples
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
edited Dec 23 '18 at 14:02
Martin Sleziak
45k10123277
45k10123277
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
asked Dec 22 '18 at 15:24
Fathima JasmineFathima Jasmine
213
213
closed as off-topic by Namaste, Holo, Cesareo, Saad, Did Dec 25 '18 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Cesareo, Saad, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Holo, Cesareo, Saad, Did Dec 25 '18 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Cesareo, Saad, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Think about sums and squares of odds and evens.
$endgroup$
– Randall
Dec 22 '18 at 15:24
$begingroup$
Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic
$endgroup$
– Fathima Jasmine
Dec 22 '18 at 15:31
$begingroup$
All even: 6, 8, 10.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:55
add a comment |
3
$begingroup$
Think about sums and squares of odds and evens.
$endgroup$
– Randall
Dec 22 '18 at 15:24
$begingroup$
Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic
$endgroup$
– Fathima Jasmine
Dec 22 '18 at 15:31
$begingroup$
All even: 6, 8, 10.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:55
3
3
$begingroup$
Think about sums and squares of odds and evens.
$endgroup$
– Randall
Dec 22 '18 at 15:24
$begingroup$
Think about sums and squares of odds and evens.
$endgroup$
– Randall
Dec 22 '18 at 15:24
$begingroup$
Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic
$endgroup$
– Fathima Jasmine
Dec 22 '18 at 15:31
$begingroup$
Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic
$endgroup$
– Fathima Jasmine
Dec 22 '18 at 15:31
$begingroup$
All even: 6, 8, 10.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:55
$begingroup$
All even: 6, 8, 10.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:55
add a comment |
3 Answers
3
active
oldest
votes
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Suppose the numbers are $a,b,c$. Suppose $a^2+b^2=c^2$. The order doesn't matter anyway. Since $a$ and $b$ are odd, $a^2+b^2$ will be even. But, $c^2$ is odd. Hence, no such triplet can exist.
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
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add a comment |
$begingroup$
Think about
$a^2b^2(a^2+b^2)$ mod $2$
$a^2b^2(a^2+b^2+2ab)$ mod $2$
$a^2b^2(a+b)^2$ mod $2$
If both $a$ and $b$ are such that $a equiv 1 text{ mod } 2$ and $b equiv 1 text{ mod } 2$ then $a+b equiv 0 text{ mod } 2$ hence abc is divisible by 2.
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add a comment |
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Use that $$a^2+b^2=c^2iff(lambda^2-mu^2)^2+(2lambdamu)^2=(lambda^2+mu^2)^2$$
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose the numbers are $a,b,c$. Suppose $a^2+b^2=c^2$. The order doesn't matter anyway. Since $a$ and $b$ are odd, $a^2+b^2$ will be even. But, $c^2$ is odd. Hence, no such triplet can exist.
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
Suppose the numbers are $a,b,c$. Suppose $a^2+b^2=c^2$. The order doesn't matter anyway. Since $a$ and $b$ are odd, $a^2+b^2$ will be even. But, $c^2$ is odd. Hence, no such triplet can exist.
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
Suppose the numbers are $a,b,c$. Suppose $a^2+b^2=c^2$. The order doesn't matter anyway. Since $a$ and $b$ are odd, $a^2+b^2$ will be even. But, $c^2$ is odd. Hence, no such triplet can exist.
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
$endgroup$
Suppose the numbers are $a,b,c$. Suppose $a^2+b^2=c^2$. The order doesn't matter anyway. Since $a$ and $b$ are odd, $a^2+b^2$ will be even. But, $c^2$ is odd. Hence, no such triplet can exist.
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 15:31
Ankit KumarAnkit Kumar
1,542221
1,542221
add a comment |
add a comment |
$begingroup$
Think about
$a^2b^2(a^2+b^2)$ mod $2$
$a^2b^2(a^2+b^2+2ab)$ mod $2$
$a^2b^2(a+b)^2$ mod $2$
If both $a$ and $b$ are such that $a equiv 1 text{ mod } 2$ and $b equiv 1 text{ mod } 2$ then $a+b equiv 0 text{ mod } 2$ hence abc is divisible by 2.
$endgroup$
add a comment |
$begingroup$
Think about
$a^2b^2(a^2+b^2)$ mod $2$
$a^2b^2(a^2+b^2+2ab)$ mod $2$
$a^2b^2(a+b)^2$ mod $2$
If both $a$ and $b$ are such that $a equiv 1 text{ mod } 2$ and $b equiv 1 text{ mod } 2$ then $a+b equiv 0 text{ mod } 2$ hence abc is divisible by 2.
$endgroup$
add a comment |
$begingroup$
Think about
$a^2b^2(a^2+b^2)$ mod $2$
$a^2b^2(a^2+b^2+2ab)$ mod $2$
$a^2b^2(a+b)^2$ mod $2$
If both $a$ and $b$ are such that $a equiv 1 text{ mod } 2$ and $b equiv 1 text{ mod } 2$ then $a+b equiv 0 text{ mod } 2$ hence abc is divisible by 2.
$endgroup$
Think about
$a^2b^2(a^2+b^2)$ mod $2$
$a^2b^2(a^2+b^2+2ab)$ mod $2$
$a^2b^2(a+b)^2$ mod $2$
If both $a$ and $b$ are such that $a equiv 1 text{ mod } 2$ and $b equiv 1 text{ mod } 2$ then $a+b equiv 0 text{ mod } 2$ hence abc is divisible by 2.
answered Dec 22 '18 at 15:40
user408906
add a comment |
add a comment |
$begingroup$
Use that $$a^2+b^2=c^2iff(lambda^2-mu^2)^2+(2lambdamu)^2=(lambda^2+mu^2)^2$$
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
$endgroup$
add a comment |
$begingroup$
Use that $$a^2+b^2=c^2iff(lambda^2-mu^2)^2+(2lambdamu)^2=(lambda^2+mu^2)^2$$
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
$endgroup$
add a comment |
$begingroup$
Use that $$a^2+b^2=c^2iff(lambda^2-mu^2)^2+(2lambdamu)^2=(lambda^2+mu^2)^2$$
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
$endgroup$
Use that $$a^2+b^2=c^2iff(lambda^2-mu^2)^2+(2lambdamu)^2=(lambda^2+mu^2)^2$$
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
answered Dec 23 '18 at 12:45
Rhys HughesRhys Hughes
7,0851630
7,0851630
add a comment |
add a comment |
3
$begingroup$
Think about sums and squares of odds and evens.
$endgroup$
– Randall
Dec 22 '18 at 15:24
$begingroup$
Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic
$endgroup$
– Fathima Jasmine
Dec 22 '18 at 15:31
$begingroup$
All even: 6, 8, 10.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:55