Is it true that every pythagorean triples is of the form $(a^2-b^2)^2+(2ab)^2=(a^2+b^2)$?
1
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For example, the Pythagorean triples $(3,4,5)$ is $(2^2-1^2)^2+(2times2times1)^2=(2^2+1^2)^2$.
number-theory
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
asked Dec 8 '18 at 12:37
user42493user42493
1837
$endgroup$
add a comment |
1
$begingroup$
For example, the Pythagorean triples $(3,4,5)$ is $(2^2-1^2)^2+(2times2times1)^2=(2^2+1^2)^2$.
number-theory
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
asked Dec 8 '18 at 12:37
user42493user42493
1837
$endgroup$
$begingroup$
See en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
$endgroup$
– lhf
Dec 8 '18 at 12:53
add a comment |
1
1
1
$begingroup$
For example, the Pythagorean triples $(3,4,5)$ is $(2^2-1^2)^2+(2times2times1)^2=(2^2+1^2)^2$.
number-theory
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
asked Dec 8 '18 at 12:37
user42493user42493
1837
$endgroup$
For example, the Pythagorean triples $(3,4,5)$ is $(2^2-1^2)^2+(2times2times1)^2=(2^2+1^2)^2$.
number-theory
number-theory
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
asked Dec 8 '18 at 12:37
user42493user42493
1837
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
asked Dec 8 '18 at 12:37
user42493user42493
1837
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
edited Dec 8 '18 at 12:54
InsideOut
4,94131033
4,94131033
asked Dec 8 '18 at 12:37
user42493user42493
1837
asked Dec 8 '18 at 12:37
user42493user42493
1837
asked Dec 8 '18 at 12:37
user42493user42493
1837
1837
$begingroup$
See en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
$endgroup$
– lhf
Dec 8 '18 at 12:53
add a comment |
$begingroup$
See en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
$endgroup$
– lhf
Dec 8 '18 at 12:53
$begingroup$
See en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
$endgroup$
– lhf
Dec 8 '18 at 12:53
$begingroup$
See en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
$endgroup$
– lhf
Dec 8 '18 at 12:53
add a comment |
1 Answer
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6
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Primitive Pythagorean triples must have that form. But there can also be triples with a common factor that prevents the hypoteneuse from being a sum of squares (like a factor of 3 in 9-12-15).
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
$begingroup$
Primitive Pythagorean triples must have that form. But there can also be triples with a common factor that prevents the hypoteneuse from being a sum of squares (like a factor of 3 in 9-12-15).
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
add a comment |
6
$begingroup$
Primitive Pythagorean triples must have that form. But there can also be triples with a common factor that prevents the hypoteneuse from being a sum of squares (like a factor of 3 in 9-12-15).
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
add a comment |
6
6
6
$begingroup$
Primitive Pythagorean triples must have that form. But there can also be triples with a common factor that prevents the hypoteneuse from being a sum of squares (like a factor of 3 in 9-12-15).
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
Primitive Pythagorean triples must have that form. But there can also be triples with a common factor that prevents the hypoteneuse from being a sum of squares (like a factor of 3 in 9-12-15).
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
answered Dec 8 '18 at 12:41
Oscar LanziOscar Lanzi
12.4k12036
12.4k12036
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
add a comment |
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
Why does having common factor greater than 1 implies the hypotenuse is not sum of two squares?
$endgroup$
– user42493
Dec 8 '18 at 12:48
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
There could be a common factor that prevents it, specifically one that is itself not a square or the sum of two squares. There is no claim here that all common factors have this property.
$endgroup$
– Oscar Lanzi
Dec 8 '18 at 12:59
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
$begingroup$
Hint-If $7|a^2+b^2$ then $7|a$ and $7|b$
$endgroup$
– Rakesh Bhatt
Dec 8 '18 at 13:00
add a comment |
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$begingroup$
See en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
$endgroup$
– lhf
Dec 8 '18 at 12:53