Another proof for an infinite number of Pythagorean triples
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I’m not sure if this has been mentioned before (and I truly apologize if someone thought about it already) , but I tried to adopt a geometrical approach for the proof for an infinite number of Pythagorean triplets. Please read it once and please point out any mistakes .
Any Pythagorean triplet can be expressed in the form of $x^2+ y^2=r^2$ (r being an integer). Keeping $r$ constant , we can easily conclude that the equation is that of a circle. Rearranging the equation gives $y= sqrt{r^2-x^2}$. (considering only natural numbers) . Now the domain of $x$ is $-r<x<r$ . Considering the base case , $r=1$ there is at least one integer pair that satisfies its equation , it being $(0,1)$. For $r>1$ , any integral value of $x$ gives a corresponding integral value of $y$, while the condition $-r<x<r$ still holds true. As there are infinite number of natural numbers $r$ , the result follows.
I would really like to know if I’ve missed on something so please guide me . Thank you !
number-theory pythagorean-triples
asked Nov 12 at 16:46
Aditi
729314
|
show 5 more comments
up vote
0
down vote
favorite
I’m not sure if this has been mentioned before (and I truly apologize if someone thought about it already) , but I tried to adopt a geometrical approach for the proof for an infinite number of Pythagorean triplets. Please read it once and please point out any mistakes .
Any Pythagorean triplet can be expressed in the form of $x^2+ y^2=r^2$ (r being an integer). Keeping $r$ constant , we can easily conclude that the equation is that of a circle. Rearranging the equation gives $y= sqrt{r^2-x^2}$. (considering only natural numbers) . Now the domain of $x$ is $-r<x<r$ . Considering the base case , $r=1$ there is at least one integer pair that satisfies its equation , it being $(0,1)$. For $r>1$ , any integral value of $x$ gives a corresponding integral value of $y$, while the condition $-r<x<r$ still holds true. As there are infinite number of natural numbers $r$ , the result follows.
I would really like to know if I’ve missed on something so please guide me . Thank you !
number-theory pythagorean-triples
asked Nov 12 at 16:46
Aditi
729314
2
"As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way.
– Rushabh Mehta
Nov 12 at 16:50
1
Please check: youtube.com/watch?v=QJYmyhnaaek
– Raptor
Nov 12 at 16:55
1
"For $r>1$, any integral value of $x$ gives [via $y=sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$.
– Blue
Nov 12 at 17:10
1
@Blue except of course the trivial case of $x=0$.
– Rushabh Mehta
Nov 12 at 17:19
1
@Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach.
– Rushabh Mehta
Nov 12 at 17:21
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I’m not sure if this has been mentioned before (and I truly apologize if someone thought about it already) , but I tried to adopt a geometrical approach for the proof for an infinite number of Pythagorean triplets. Please read it once and please point out any mistakes .
Any Pythagorean triplet can be expressed in the form of $x^2+ y^2=r^2$ (r being an integer). Keeping $r$ constant , we can easily conclude that the equation is that of a circle. Rearranging the equation gives $y= sqrt{r^2-x^2}$. (considering only natural numbers) . Now the domain of $x$ is $-r<x<r$ . Considering the base case , $r=1$ there is at least one integer pair that satisfies its equation , it being $(0,1)$. For $r>1$ , any integral value of $x$ gives a corresponding integral value of $y$, while the condition $-r<x<r$ still holds true. As there are infinite number of natural numbers $r$ , the result follows.
I would really like to know if I’ve missed on something so please guide me . Thank you !
number-theory pythagorean-triples
asked Nov 12 at 16:46
Aditi
729314
I’m not sure if this has been mentioned before (and I truly apologize if someone thought about it already) , but I tried to adopt a geometrical approach for the proof for an infinite number of Pythagorean triplets. Please read it once and please point out any mistakes .
Any Pythagorean triplet can be expressed in the form of $x^2+ y^2=r^2$ (r being an integer). Keeping $r$ constant , we can easily conclude that the equation is that of a circle. Rearranging the equation gives $y= sqrt{r^2-x^2}$. (considering only natural numbers) . Now the domain of $x$ is $-r<x<r$ . Considering the base case , $r=1$ there is at least one integer pair that satisfies its equation , it being $(0,1)$. For $r>1$ , any integral value of $x$ gives a corresponding integral value of $y$, while the condition $-r<x<r$ still holds true. As there are infinite number of natural numbers $r$ , the result follows.
I would really like to know if I’ve missed on something so please guide me . Thank you !
number-theory pythagorean-triples
number-theory pythagorean-triples
asked Nov 12 at 16:46
Aditi
729314
asked Nov 12 at 16:46
Aditi
729314
edited Nov 12 at 16:58
asked Nov 12 at 16:46
Aditi
729314
asked Nov 12 at 16:46
Aditi
729314
asked Nov 12 at 16:46
Aditi
729314
729314
2
"As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way.
– Rushabh Mehta
Nov 12 at 16:50
1
Please check: youtube.com/watch?v=QJYmyhnaaek
– Raptor
Nov 12 at 16:55
1
"For $r>1$, any integral value of $x$ gives [via $y=sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$.
– Blue
Nov 12 at 17:10
1
@Blue except of course the trivial case of $x=0$.
– Rushabh Mehta
Nov 12 at 17:19
1
@Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach.
– Rushabh Mehta
Nov 12 at 17:21
|
show 5 more comments
2
"As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way.
– Rushabh Mehta
Nov 12 at 16:50
1
Please check: youtube.com/watch?v=QJYmyhnaaek
– Raptor
Nov 12 at 16:55
1
"For $r>1$, any integral value of $x$ gives [via $y=sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$.
– Blue
Nov 12 at 17:10
1
@Blue except of course the trivial case of $x=0$.
– Rushabh Mehta
Nov 12 at 17:19
1
@Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach.
– Rushabh Mehta
Nov 12 at 17:21
2
2
"As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way.
– Rushabh Mehta
Nov 12 at 16:50
"As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way.
– Rushabh Mehta
Nov 12 at 16:50
1
1
Please check: youtube.com/watch?v=QJYmyhnaaek
– Raptor
Nov 12 at 16:55
Please check: youtube.com/watch?v=QJYmyhnaaek
– Raptor
Nov 12 at 16:55
1
1
"For $r>1$, any integral value of $x$ gives [via $y=sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$.
– Blue
Nov 12 at 17:10
"For $r>1$, any integral value of $x$ gives [via $y=sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$.
– Blue
Nov 12 at 17:10
1
1
@Blue except of course the trivial case of $x=0$.
– Rushabh Mehta
Nov 12 at 17:19
@Blue except of course the trivial case of $x=0$.
– Rushabh Mehta
Nov 12 at 17:19
1
1
@Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach.
– Rushabh Mehta
Nov 12 at 17:21
@Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach.
– Rushabh Mehta
Nov 12 at 17:21
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your statement is correct but your proof is not.
For example, consider the equation $$y= sqrt {16-x^2}$$
You have claimed that for any integral value of $x$ you get an integral value of $y$
For $x= 1, 2, 3$ you get $y= sqrt {15} , sqrt {12}, sqrt {7} $ and none of these numbers are integers.
The only integers that we get out of that are $0$ and $4$ which are trivial solutions to $ x^2+y^2=16$
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
1
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your statement is correct but your proof is not.
For example, consider the equation $$y= sqrt {16-x^2}$$
You have claimed that for any integral value of $x$ you get an integral value of $y$
For $x= 1, 2, 3$ you get $y= sqrt {15} , sqrt {12}, sqrt {7} $ and none of these numbers are integers.
The only integers that we get out of that are $0$ and $4$ which are trivial solutions to $ x^2+y^2=16$
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
1
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
add a comment |
up vote
1
down vote
accepted
Your statement is correct but your proof is not.
For example, consider the equation $$y= sqrt {16-x^2}$$
You have claimed that for any integral value of $x$ you get an integral value of $y$
For $x= 1, 2, 3$ you get $y= sqrt {15} , sqrt {12}, sqrt {7} $ and none of these numbers are integers.
The only integers that we get out of that are $0$ and $4$ which are trivial solutions to $ x^2+y^2=16$
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
1
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your statement is correct but your proof is not.
For example, consider the equation $$y= sqrt {16-x^2}$$
You have claimed that for any integral value of $x$ you get an integral value of $y$
For $x= 1, 2, 3$ you get $y= sqrt {15} , sqrt {12}, sqrt {7} $ and none of these numbers are integers.
The only integers that we get out of that are $0$ and $4$ which are trivial solutions to $ x^2+y^2=16$
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
Your statement is correct but your proof is not.
For example, consider the equation $$y= sqrt {16-x^2}$$
You have claimed that for any integral value of $x$ you get an integral value of $y$
For $x= 1, 2, 3$ you get $y= sqrt {15} , sqrt {12}, sqrt {7} $ and none of these numbers are integers.
The only integers that we get out of that are $0$ and $4$ which are trivial solutions to $ x^2+y^2=16$
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
answered Nov 12 at 17:10
Mohammad Riazi-Kermani
39.7k41957
39.7k41957
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
1
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
add a comment |
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
1
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that?
– Aditi
Nov 12 at 17:13
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake !
– Aditi
Nov 12 at 17:20
1
1
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
@Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work.
– Mohammad Riazi-Kermani
Nov 12 at 17:30
add a comment |
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2
"As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way.
– Rushabh Mehta
Nov 12 at 16:50
1
Please check: youtube.com/watch?v=QJYmyhnaaek
– Raptor
Nov 12 at 16:55
1
"For $r>1$, any integral value of $x$ gives [via $y=sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$.
– Blue
Nov 12 at 17:10
1
@Blue except of course the trivial case of $x=0$.
– Rushabh Mehta
Nov 12 at 17:19
1
@Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach.
– Rushabh Mehta
Nov 12 at 17:21