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How to obtain the bounds for a and b?
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If $a,b,c$ are Pythagorean triplets. Provided $a+b+c = s$ and $a<b<c$. We can conclude that $alefracs-33$ and $b<fracs-a2$.
From the given conditions we can obtain:
$a^2 + b^2 = (s-a-b)^2$.
I am unable to obtain those bounds for $a$ and $b$.
I tried $2a^2 < a^2 + b^2 = (s-a-b)^2 < (s-2a)^2$
which gives
$2a^2 > s^2 - 2as$
I don't think this can provide me those bounds.
number-theory elementary-number-theory pythagorean-triples
asked Aug 1 at 9:59
reego
787416
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If $a,b,c$ are Pythagorean triplets. Provided $a+b+c = s$ and $a<b<c$. We can conclude that $alefracs-33$ and $b<fracs-a2$.
From the given conditions we can obtain:
$a^2 + b^2 = (s-a-b)^2$.
I am unable to obtain those bounds for $a$ and $b$.
I tried $2a^2 < a^2 + b^2 = (s-a-b)^2 < (s-2a)^2$
which gives
$2a^2 > s^2 - 2as$
I don't think this can provide me those bounds.
number-theory elementary-number-theory pythagorean-triples
asked Aug 1 at 9:59
reego
787416
2
For what stands $s$ here?
– Dr. Sonnhard Graubner
Aug 1 at 10:03
@Dr.SonnhardGraubner I made a mistake.
– reego
Aug 1 at 10:06
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $a,b,c$ are Pythagorean triplets. Provided $a+b+c = s$ and $a<b<c$. We can conclude that $alefracs-33$ and $b<fracs-a2$.
From the given conditions we can obtain:
$a^2 + b^2 = (s-a-b)^2$.
I am unable to obtain those bounds for $a$ and $b$.
I tried $2a^2 < a^2 + b^2 = (s-a-b)^2 < (s-2a)^2$
which gives
$2a^2 > s^2 - 2as$
I don't think this can provide me those bounds.
number-theory elementary-number-theory pythagorean-triples
asked Aug 1 at 9:59
reego
787416
If $a,b,c$ are Pythagorean triplets. Provided $a+b+c = s$ and $a<b<c$. We can conclude that $alefracs-33$ and $b<fracs-a2$.
From the given conditions we can obtain:
$a^2 + b^2 = (s-a-b)^2$.
I am unable to obtain those bounds for $a$ and $b$.
I tried $2a^2 < a^2 + b^2 = (s-a-b)^2 < (s-2a)^2$
which gives
$2a^2 > s^2 - 2as$
I don't think this can provide me those bounds.
number-theory elementary-number-theory pythagorean-triples
asked Aug 1 at 9:59
reego
787416
edited Aug 1 at 10:05
asked Aug 1 at 9:59
reego
787416
asked Aug 1 at 9:59
reego
787416
asked Aug 1 at 9:59
reego
787416
787416
2
For what stands $s$ here?
– Dr. Sonnhard Graubner
Aug 1 at 10:03
@Dr.SonnhardGraubner I made a mistake.
– reego
Aug 1 at 10:06
add a comment |Â
2
For what stands $s$ here?
– Dr. Sonnhard Graubner
Aug 1 at 10:03
@Dr.SonnhardGraubner I made a mistake.
– reego
Aug 1 at 10:06
2
2
For what stands $s$ here?
– Dr. Sonnhard Graubner
Aug 1 at 10:03
For what stands $s$ here?
– Dr. Sonnhard Graubner
Aug 1 at 10:03
@Dr.SonnhardGraubner I made a mistake.
– reego
Aug 1 at 10:06
@Dr.SonnhardGraubner I made a mistake.
– reego
Aug 1 at 10:06
add a comment |Â
2 Answers
2
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Are you asking for a proof of $$aleq dfracs-33text and b<dfracs-a2,?$$ If so, note that $bgeq a+1$ and $cgeq b+1geq a+2$. Therefore,
$$s=a+b+cgeq a+(a+1)+(a+2),,$$
and the first claim is established. (Note that the only equality case for this inequality is $(a,b,c)=(3,4,5)$.)
The second claim follows from $b<c$, namely,
$$s-a=b+c>b+b=2b,.$$
We can obtain a nicer-looking (in my opinion) bound $$bleq dfracs-a-12$$ in this manner too (with the equality case $(a,b,c)=left(2k+1,2k^2+2k,2k^2+2k+1right)$ where $kinmathbbZ_>0$).
answered Aug 1 at 10:11
Batominovski
22.7k22776
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up vote
0
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It's known that if $(a,b,c)$ is a Pythagorean triples then there are natural $m$ and $n$ for which $m>n$ and
$$a,b,c=m^2-n^2,2mn,m^2+n^2$$
Thus, $m(m+n)=500$ and number of cases.
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Are you asking for a proof of $$aleq dfracs-33text and b<dfracs-a2,?$$ If so, note that $bgeq a+1$ and $cgeq b+1geq a+2$. Therefore,
$$s=a+b+cgeq a+(a+1)+(a+2),,$$
and the first claim is established. (Note that the only equality case for this inequality is $(a,b,c)=(3,4,5)$.)
The second claim follows from $b<c$, namely,
$$s-a=b+c>b+b=2b,.$$
We can obtain a nicer-looking (in my opinion) bound $$bleq dfracs-a-12$$ in this manner too (with the equality case $(a,b,c)=left(2k+1,2k^2+2k,2k^2+2k+1right)$ where $kinmathbbZ_>0$).
answered Aug 1 at 10:11
Batominovski
22.7k22776
add a comment |Â
up vote
1
down vote
accepted
Are you asking for a proof of $$aleq dfracs-33text and b<dfracs-a2,?$$ If so, note that $bgeq a+1$ and $cgeq b+1geq a+2$. Therefore,
$$s=a+b+cgeq a+(a+1)+(a+2),,$$
and the first claim is established. (Note that the only equality case for this inequality is $(a,b,c)=(3,4,5)$.)
The second claim follows from $b<c$, namely,
$$s-a=b+c>b+b=2b,.$$
We can obtain a nicer-looking (in my opinion) bound $$bleq dfracs-a-12$$ in this manner too (with the equality case $(a,b,c)=left(2k+1,2k^2+2k,2k^2+2k+1right)$ where $kinmathbbZ_>0$).
answered Aug 1 at 10:11
Batominovski
22.7k22776
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Are you asking for a proof of $$aleq dfracs-33text and b<dfracs-a2,?$$ If so, note that $bgeq a+1$ and $cgeq b+1geq a+2$. Therefore,
$$s=a+b+cgeq a+(a+1)+(a+2),,$$
and the first claim is established. (Note that the only equality case for this inequality is $(a,b,c)=(3,4,5)$.)
The second claim follows from $b<c$, namely,
$$s-a=b+c>b+b=2b,.$$
We can obtain a nicer-looking (in my opinion) bound $$bleq dfracs-a-12$$ in this manner too (with the equality case $(a,b,c)=left(2k+1,2k^2+2k,2k^2+2k+1right)$ where $kinmathbbZ_>0$).
answered Aug 1 at 10:11
Batominovski
22.7k22776
Are you asking for a proof of $$aleq dfracs-33text and b<dfracs-a2,?$$ If so, note that $bgeq a+1$ and $cgeq b+1geq a+2$. Therefore,
$$s=a+b+cgeq a+(a+1)+(a+2),,$$
and the first claim is established. (Note that the only equality case for this inequality is $(a,b,c)=(3,4,5)$.)
The second claim follows from $b<c$, namely,
$$s-a=b+c>b+b=2b,.$$
We can obtain a nicer-looking (in my opinion) bound $$bleq dfracs-a-12$$ in this manner too (with the equality case $(a,b,c)=left(2k+1,2k^2+2k,2k^2+2k+1right)$ where $kinmathbbZ_>0$).
answered Aug 1 at 10:11
Batominovski
22.7k22776
edited Aug 1 at 10:16
answered Aug 1 at 10:11
Batominovski
22.7k22776
answered Aug 1 at 10:11
Batominovski
22.7k22776
answered Aug 1 at 10:11
Batominovski
22.7k22776
22.7k22776
add a comment |Â
add a comment |Â
up vote
0
down vote
It's known that if $(a,b,c)$ is a Pythagorean triples then there are natural $m$ and $n$ for which $m>n$ and
$$a,b,c=m^2-n^2,2mn,m^2+n^2$$
Thus, $m(m+n)=500$ and number of cases.
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
add a comment |Â
up vote
0
down vote
It's known that if $(a,b,c)$ is a Pythagorean triples then there are natural $m$ and $n$ for which $m>n$ and
$$a,b,c=m^2-n^2,2mn,m^2+n^2$$
Thus, $m(m+n)=500$ and number of cases.
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It's known that if $(a,b,c)$ is a Pythagorean triples then there are natural $m$ and $n$ for which $m>n$ and
$$a,b,c=m^2-n^2,2mn,m^2+n^2$$
Thus, $m(m+n)=500$ and number of cases.
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
It's known that if $(a,b,c)$ is a Pythagorean triples then there are natural $m$ and $n$ for which $m>n$ and
$$a,b,c=m^2-n^2,2mn,m^2+n^2$$
Thus, $m(m+n)=500$ and number of cases.
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
answered Aug 1 at 10:09
Michael Rozenberg
87.4k1577179
87.4k1577179
add a comment |Â
add a comment |Â
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2
For what stands $s$ here?
– Dr. Sonnhard Graubner
Aug 1 at 10:03
@Dr.SonnhardGraubner I made a mistake.
– reego
Aug 1 at 10:06