Pythagorean Triplets with “Bounds”
up vote
3
down vote
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I am interested in the algebraic/geometric way of finding the pythagorean triplets such that
$$a^2 + b^2 = c^2$$
$$a + b + c = 1000$$
I do the obvious
$$a + b = 1000 - (a^2 + b^2)^{1/2}$$
$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$
$$2a + 2b - frac{2ab}{1000} = 1000$$
$$a + b -frac{ab}{1000} = 500$$
I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing
$$a + b = 500 + frac{ab}{1000} = 1000 - c$$
But this is going backwards
algebra-precalculus
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
add a comment |
up vote
3
down vote
favorite
I am interested in the algebraic/geometric way of finding the pythagorean triplets such that
$$a^2 + b^2 = c^2$$
$$a + b + c = 1000$$
I do the obvious
$$a + b = 1000 - (a^2 + b^2)^{1/2}$$
$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$
$$2a + 2b - frac{2ab}{1000} = 1000$$
$$a + b -frac{ab}{1000} = 500$$
I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing
$$a + b = 500 + frac{ab}{1000} = 1000 - c$$
But this is going backwards
algebra-precalculus
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am interested in the algebraic/geometric way of finding the pythagorean triplets such that
$$a^2 + b^2 = c^2$$
$$a + b + c = 1000$$
I do the obvious
$$a + b = 1000 - (a^2 + b^2)^{1/2}$$
$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$
$$2a + 2b - frac{2ab}{1000} = 1000$$
$$a + b -frac{ab}{1000} = 500$$
I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing
$$a + b = 500 + frac{ab}{1000} = 1000 - c$$
But this is going backwards
algebra-precalculus
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
I am interested in the algebraic/geometric way of finding the pythagorean triplets such that
$$a^2 + b^2 = c^2$$
$$a + b + c = 1000$$
I do the obvious
$$a + b = 1000 - (a^2 + b^2)^{1/2}$$
$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$
$$2a + 2b - frac{2ab}{1000} = 1000$$
$$a + b -frac{ab}{1000} = 500$$
I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing
$$a + b = 500 + frac{ab}{1000} = 1000 - c$$
But this is going backwards
algebra-precalculus
algebra-precalculus
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
edited Aug 7 '12 at 16:41
Micah
29.5k1363104
29.5k1363104
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
asked Aug 7 '12 at 16:33
Cactus BAMF
5441920
5441920
Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01
add a comment |
Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01
Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38
Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The following is copy and pasted directly from Yahoo Answers
All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.
You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.
$m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
and
$m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.
The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
Thus, $m=20$, $n=5$, giving the answer required:
$m^2+n^2$ $= 425$;
$m^2-n^2 = 375$;
$2mn = 200$.
edited Nov 19 at 20:38
Ben Brian
1034
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
add a comment |
up vote
1
down vote
We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.
So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$
=>$kp(p+q)=500$
Clearly, k can be any divisor of 500.
If k=1,
If p=1, p+q=500=> q = 499,
if p=2, p+q=250=> q = 248 and so on.
If k=2,$p(p+q)=250$
If k=5,$p(p+q)=100$
If k=10, $p(p+q)=50$ and so on.
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
add a comment |
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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
The following is copy and pasted directly from Yahoo Answers
All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.
You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.
$m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
and
$m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.
The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
Thus, $m=20$, $n=5$, giving the answer required:
$m^2+n^2$ $= 425$;
$m^2-n^2 = 375$;
$2mn = 200$.
edited Nov 19 at 20:38
Ben Brian
1034
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
add a comment |
up vote
1
down vote
accepted
The following is copy and pasted directly from Yahoo Answers
All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.
You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.
$m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
and
$m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.
The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
Thus, $m=20$, $n=5$, giving the answer required:
$m^2+n^2$ $= 425$;
$m^2-n^2 = 375$;
$2mn = 200$.
edited Nov 19 at 20:38
Ben Brian
1034
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The following is copy and pasted directly from Yahoo Answers
All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.
You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.
$m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
and
$m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.
The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
Thus, $m=20$, $n=5$, giving the answer required:
$m^2+n^2$ $= 425$;
$m^2-n^2 = 375$;
$2mn = 200$.
edited Nov 19 at 20:38
Ben Brian
1034
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
The following is copy and pasted directly from Yahoo Answers
All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.
You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.
$m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
and
$m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.
The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
Thus, $m=20$, $n=5$, giving the answer required:
$m^2+n^2$ $= 425$;
$m^2-n^2 = 375$;
$2mn = 200$.
edited Nov 19 at 20:38
Ben Brian
1034
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
edited Nov 19 at 20:38
Ben Brian
1034
edited Nov 19 at 20:38
Ben Brian
1034
edited Nov 19 at 20:38
Ben Brian
1034
1034
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
answered Aug 7 '12 at 16:47
Sidd Singal
2,56231630
2,56231630
add a comment |
add a comment |
up vote
1
down vote
We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.
So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$
=>$kp(p+q)=500$
Clearly, k can be any divisor of 500.
If k=1,
If p=1, p+q=500=> q = 499,
if p=2, p+q=250=> q = 248 and so on.
If k=2,$p(p+q)=250$
If k=5,$p(p+q)=100$
If k=10, $p(p+q)=50$ and so on.
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
add a comment |
up vote
1
down vote
We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.
So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$
=>$kp(p+q)=500$
Clearly, k can be any divisor of 500.
If k=1,
If p=1, p+q=500=> q = 499,
if p=2, p+q=250=> q = 248 and so on.
If k=2,$p(p+q)=250$
If k=5,$p(p+q)=100$
If k=10, $p(p+q)=50$ and so on.
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
add a comment |
up vote
1
down vote
up vote
1
down vote
We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.
So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$
=>$kp(p+q)=500$
Clearly, k can be any divisor of 500.
If k=1,
If p=1, p+q=500=> q = 499,
if p=2, p+q=250=> q = 248 and so on.
If k=2,$p(p+q)=250$
If k=5,$p(p+q)=100$
If k=10, $p(p+q)=50$ and so on.
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.
So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$
=>$kp(p+q)=500$
Clearly, k can be any divisor of 500.
If k=1,
If p=1, p+q=500=> q = 499,
if p=2, p+q=250=> q = 248 and so on.
If k=2,$p(p+q)=250$
If k=5,$p(p+q)=100$
If k=10, $p(p+q)=50$ and so on.
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
edited Aug 7 '12 at 16:53
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
answered Aug 7 '12 at 16:41
lab bhattacharjee
222k15155273
222k15155273
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
add a comment |
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
– Qiaochu Yuan
Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
– lab bhattacharjee
Aug 7 '12 at 18:00
add a comment |
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Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01