Pythagorean triple
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Show that neither $1$ not $2$ can appear in any Pythagorean triple, but that every integer $kgeq3$ can appear.
Prove that for each integer $k$ there are only finitely many Pythagorean triple containing $k$.
Help me to prove this...
Thank you.
number-theory pythagorean-triples
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favorite
Show that neither $1$ not $2$ can appear in any Pythagorean triple, but that every integer $kgeq3$ can appear.
Prove that for each integer $k$ there are only finitely many Pythagorean triple containing $k$.
Help me to prove this...
Thank you.
number-theory pythagorean-triples
1
Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you.
– Kyky
Nov 17 at 1:55
I have no idea......
– Nithish Kumar R
Nov 17 at 1:59
If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers.
– rafa11111
Nov 17 at 2:06
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up vote
0
down vote
favorite
Show that neither $1$ not $2$ can appear in any Pythagorean triple, but that every integer $kgeq3$ can appear.
Prove that for each integer $k$ there are only finitely many Pythagorean triple containing $k$.
Help me to prove this...
Thank you.
number-theory pythagorean-triples
Show that neither $1$ not $2$ can appear in any Pythagorean triple, but that every integer $kgeq3$ can appear.
Prove that for each integer $k$ there are only finitely many Pythagorean triple containing $k$.
Help me to prove this...
Thank you.
number-theory pythagorean-triples
number-theory pythagorean-triples
edited Nov 17 at 1:56
Tianlalu
2,8811935
2,8811935
asked Nov 17 at 1:52
Nithish Kumar R
91
91
1
Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you.
– Kyky
Nov 17 at 1:55
I have no idea......
– Nithish Kumar R
Nov 17 at 1:59
If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers.
– rafa11111
Nov 17 at 2:06
add a comment |
1
Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you.
– Kyky
Nov 17 at 1:55
I have no idea......
– Nithish Kumar R
Nov 17 at 1:59
If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers.
– rafa11111
Nov 17 at 2:06
1
1
Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you.
– Kyky
Nov 17 at 1:55
Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you.
– Kyky
Nov 17 at 1:55
I have no idea......
– Nithish Kumar R
Nov 17 at 1:59
I have no idea......
– Nithish Kumar R
Nov 17 at 1:59
If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers.
– rafa11111
Nov 17 at 2:06
If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers.
– rafa11111
Nov 17 at 2:06
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3 Answers
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$1^2$ and $2^2$ are not sums of two positive integers squares so for any Pythagorean
triple $(x, y, z)$ we must have $z ge 3$.
Now $x^2 < x^2 + 1 < (x + 1)^2$ for any $x in N$, so $x^2 + 1^2$ is not a perfect square which gives $1 in { x, y, z}$. i.e. $x ne 1, y ne 1$, and so $x ge 2, y ge 2$. If $y = 2$, then $x^2 < x^2 +y^2 = x^2 + 4 < (x+ 1)^2$
, since $x ge 2$, and thus $x^2 +y^2$ is not a perfect square, so $y ne 2$. Similarly $x ne 2$. Thus ${x, y, z} ⊆ {3, 4, 5, 6, · · · }$.
Let $k ∈ N, k ge 3$ any. If $k$ is odd, then $k^2 = 2l + 1$ with $l ge 4$ and $l^2 + k^2 = (l + 1)^2$;
if $k$ is even, then $k^2 = 4l$ with $l ge 4$ and $(l − 1)^2 + k^2 = (l + 1)^2$
Thus for any integer $k ge 3$, we have a Pythagorean triple $(x, y, z)$ such that $k ∈ {x, y, z}$
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
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0
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Since the formula $a^2+b^2=c^2$ can be changed into $a^2=c^2-b^2$, $a^2$ needs to mean the difference between $2$ squares. You will find that $(a+1)^2-a^2$ is equal to $a^2-2a+1-a^2=2a+1$ and $(a+2)^2-a^2=a^2+4a+4-a^2=4a+4$. The former must be odd and the latter must be even. These are obtained by unpacking the bracket and simplifying. This is the maximum values of $c^2$ and $b^2$ of which $a^2=c^2-b^2$ stand true as if there were any larger then $c^2$ and $b^2$ are fractional. For $1$, the largest values of $c^2$ and $b^2$ is $0^2$ and $1^2$ respectively, but $0^2$ can't be part of a side. Again, $2^2$ is the difference between $0^2$ and $2^2$, but $0^2$ can't be used.
On to the second question. We have already proven that the above formulae show that there is a limit as to the values of $b^2$ and $c^2$, and if there are infinite possible values of $b^2$ and $c^2$ then some of the must be non-integer. So by proof of contradiction for any real integer $k$, there can only be a finite amount of Pythagorean triples.
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We have $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2.$
(1). The sequence $(2^2-1^2, 3^2-2^2, 4^2-3^2,...)$ of differences of successive squares is the sequence $(3,5,7,...)$ of odd numbers $>1$. So if $m $ is odd and $mgeq 3$ then we can find $b$ such that $(b+1)^2-b^2=m.$ That is, $2b+1=m.$ So let $b=(m-1)/2$ and $a=b+1.$ Then we have $$(a^2+b^2)^2=(a^2-b^2)+(2ab)^2=m^2+(2ab)^2.$$This does not work for $m=1$ as the member $2ab$ of the triplet would be $0.$
(2). If $m$ is even and $mgeq 4$, let $a=m/2$ and $b=1.$ Then we have $$(a^2+b^2)^2 =(a^2-b^2)^2+(2ab)^2=(a^2-b^2)+m^2.$$This does not work for $m=2$ as the member $a^2-b^2$ of the triplet would be $0$.
(3). If $x,y,z$ are positive integers with $x^2+y^2=z^2$ then
(3-i). It is easy to confirm that $1ne zne 2.$
(3-ii). We have $y<z$ so $ yleq z-1$ so $x^2=z^2-y^2geq z^2-(z-1)^2=2z-1geq 5$, so $x>2.$ Interchanging $x,y$ in this, we also get $y^2geq 5$ so $y>2$.
BTW. The general formula for ALL Pyth. triplets is ${k(a^2-b^2),,2kab,, k(a^2+b^2)}$ for $a,b,kin Bbb N$ with $a>b$.
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$1^2$ and $2^2$ are not sums of two positive integers squares so for any Pythagorean
triple $(x, y, z)$ we must have $z ge 3$.
Now $x^2 < x^2 + 1 < (x + 1)^2$ for any $x in N$, so $x^2 + 1^2$ is not a perfect square which gives $1 in { x, y, z}$. i.e. $x ne 1, y ne 1$, and so $x ge 2, y ge 2$. If $y = 2$, then $x^2 < x^2 +y^2 = x^2 + 4 < (x+ 1)^2$
, since $x ge 2$, and thus $x^2 +y^2$ is not a perfect square, so $y ne 2$. Similarly $x ne 2$. Thus ${x, y, z} ⊆ {3, 4, 5, 6, · · · }$.
Let $k ∈ N, k ge 3$ any. If $k$ is odd, then $k^2 = 2l + 1$ with $l ge 4$ and $l^2 + k^2 = (l + 1)^2$;
if $k$ is even, then $k^2 = 4l$ with $l ge 4$ and $(l − 1)^2 + k^2 = (l + 1)^2$
Thus for any integer $k ge 3$, we have a Pythagorean triple $(x, y, z)$ such that $k ∈ {x, y, z}$
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
add a comment |
up vote
2
down vote
$1^2$ and $2^2$ are not sums of two positive integers squares so for any Pythagorean
triple $(x, y, z)$ we must have $z ge 3$.
Now $x^2 < x^2 + 1 < (x + 1)^2$ for any $x in N$, so $x^2 + 1^2$ is not a perfect square which gives $1 in { x, y, z}$. i.e. $x ne 1, y ne 1$, and so $x ge 2, y ge 2$. If $y = 2$, then $x^2 < x^2 +y^2 = x^2 + 4 < (x+ 1)^2$
, since $x ge 2$, and thus $x^2 +y^2$ is not a perfect square, so $y ne 2$. Similarly $x ne 2$. Thus ${x, y, z} ⊆ {3, 4, 5, 6, · · · }$.
Let $k ∈ N, k ge 3$ any. If $k$ is odd, then $k^2 = 2l + 1$ with $l ge 4$ and $l^2 + k^2 = (l + 1)^2$;
if $k$ is even, then $k^2 = 4l$ with $l ge 4$ and $(l − 1)^2 + k^2 = (l + 1)^2$
Thus for any integer $k ge 3$, we have a Pythagorean triple $(x, y, z)$ such that $k ∈ {x, y, z}$
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
add a comment |
up vote
2
down vote
up vote
2
down vote
$1^2$ and $2^2$ are not sums of two positive integers squares so for any Pythagorean
triple $(x, y, z)$ we must have $z ge 3$.
Now $x^2 < x^2 + 1 < (x + 1)^2$ for any $x in N$, so $x^2 + 1^2$ is not a perfect square which gives $1 in { x, y, z}$. i.e. $x ne 1, y ne 1$, and so $x ge 2, y ge 2$. If $y = 2$, then $x^2 < x^2 +y^2 = x^2 + 4 < (x+ 1)^2$
, since $x ge 2$, and thus $x^2 +y^2$ is not a perfect square, so $y ne 2$. Similarly $x ne 2$. Thus ${x, y, z} ⊆ {3, 4, 5, 6, · · · }$.
Let $k ∈ N, k ge 3$ any. If $k$ is odd, then $k^2 = 2l + 1$ with $l ge 4$ and $l^2 + k^2 = (l + 1)^2$;
if $k$ is even, then $k^2 = 4l$ with $l ge 4$ and $(l − 1)^2 + k^2 = (l + 1)^2$
Thus for any integer $k ge 3$, we have a Pythagorean triple $(x, y, z)$ such that $k ∈ {x, y, z}$
$1^2$ and $2^2$ are not sums of two positive integers squares so for any Pythagorean
triple $(x, y, z)$ we must have $z ge 3$.
Now $x^2 < x^2 + 1 < (x + 1)^2$ for any $x in N$, so $x^2 + 1^2$ is not a perfect square which gives $1 in { x, y, z}$. i.e. $x ne 1, y ne 1$, and so $x ge 2, y ge 2$. If $y = 2$, then $x^2 < x^2 +y^2 = x^2 + 4 < (x+ 1)^2$
, since $x ge 2$, and thus $x^2 +y^2$ is not a perfect square, so $y ne 2$. Similarly $x ne 2$. Thus ${x, y, z} ⊆ {3, 4, 5, 6, · · · }$.
Let $k ∈ N, k ge 3$ any. If $k$ is odd, then $k^2 = 2l + 1$ with $l ge 4$ and $l^2 + k^2 = (l + 1)^2$;
if $k$ is even, then $k^2 = 4l$ with $l ge 4$ and $(l − 1)^2 + k^2 = (l + 1)^2$
Thus for any integer $k ge 3$, we have a Pythagorean triple $(x, y, z)$ such that $k ∈ {x, y, z}$
answered Nov 17 at 2:06
Key Flex
7,03931229
7,03931229
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
add a comment |
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
I think you mean "are not the difference of two positive integer squares" in the first phrase.
– rafa11111
Nov 17 at 2:18
add a comment |
up vote
0
down vote
Since the formula $a^2+b^2=c^2$ can be changed into $a^2=c^2-b^2$, $a^2$ needs to mean the difference between $2$ squares. You will find that $(a+1)^2-a^2$ is equal to $a^2-2a+1-a^2=2a+1$ and $(a+2)^2-a^2=a^2+4a+4-a^2=4a+4$. The former must be odd and the latter must be even. These are obtained by unpacking the bracket and simplifying. This is the maximum values of $c^2$ and $b^2$ of which $a^2=c^2-b^2$ stand true as if there were any larger then $c^2$ and $b^2$ are fractional. For $1$, the largest values of $c^2$ and $b^2$ is $0^2$ and $1^2$ respectively, but $0^2$ can't be part of a side. Again, $2^2$ is the difference between $0^2$ and $2^2$, but $0^2$ can't be used.
On to the second question. We have already proven that the above formulae show that there is a limit as to the values of $b^2$ and $c^2$, and if there are infinite possible values of $b^2$ and $c^2$ then some of the must be non-integer. So by proof of contradiction for any real integer $k$, there can only be a finite amount of Pythagorean triples.
add a comment |
up vote
0
down vote
Since the formula $a^2+b^2=c^2$ can be changed into $a^2=c^2-b^2$, $a^2$ needs to mean the difference between $2$ squares. You will find that $(a+1)^2-a^2$ is equal to $a^2-2a+1-a^2=2a+1$ and $(a+2)^2-a^2=a^2+4a+4-a^2=4a+4$. The former must be odd and the latter must be even. These are obtained by unpacking the bracket and simplifying. This is the maximum values of $c^2$ and $b^2$ of which $a^2=c^2-b^2$ stand true as if there were any larger then $c^2$ and $b^2$ are fractional. For $1$, the largest values of $c^2$ and $b^2$ is $0^2$ and $1^2$ respectively, but $0^2$ can't be part of a side. Again, $2^2$ is the difference between $0^2$ and $2^2$, but $0^2$ can't be used.
On to the second question. We have already proven that the above formulae show that there is a limit as to the values of $b^2$ and $c^2$, and if there are infinite possible values of $b^2$ and $c^2$ then some of the must be non-integer. So by proof of contradiction for any real integer $k$, there can only be a finite amount of Pythagorean triples.
add a comment |
up vote
0
down vote
up vote
0
down vote
Since the formula $a^2+b^2=c^2$ can be changed into $a^2=c^2-b^2$, $a^2$ needs to mean the difference between $2$ squares. You will find that $(a+1)^2-a^2$ is equal to $a^2-2a+1-a^2=2a+1$ and $(a+2)^2-a^2=a^2+4a+4-a^2=4a+4$. The former must be odd and the latter must be even. These are obtained by unpacking the bracket and simplifying. This is the maximum values of $c^2$ and $b^2$ of which $a^2=c^2-b^2$ stand true as if there were any larger then $c^2$ and $b^2$ are fractional. For $1$, the largest values of $c^2$ and $b^2$ is $0^2$ and $1^2$ respectively, but $0^2$ can't be part of a side. Again, $2^2$ is the difference between $0^2$ and $2^2$, but $0^2$ can't be used.
On to the second question. We have already proven that the above formulae show that there is a limit as to the values of $b^2$ and $c^2$, and if there are infinite possible values of $b^2$ and $c^2$ then some of the must be non-integer. So by proof of contradiction for any real integer $k$, there can only be a finite amount of Pythagorean triples.
Since the formula $a^2+b^2=c^2$ can be changed into $a^2=c^2-b^2$, $a^2$ needs to mean the difference between $2$ squares. You will find that $(a+1)^2-a^2$ is equal to $a^2-2a+1-a^2=2a+1$ and $(a+2)^2-a^2=a^2+4a+4-a^2=4a+4$. The former must be odd and the latter must be even. These are obtained by unpacking the bracket and simplifying. This is the maximum values of $c^2$ and $b^2$ of which $a^2=c^2-b^2$ stand true as if there were any larger then $c^2$ and $b^2$ are fractional. For $1$, the largest values of $c^2$ and $b^2$ is $0^2$ and $1^2$ respectively, but $0^2$ can't be part of a side. Again, $2^2$ is the difference between $0^2$ and $2^2$, but $0^2$ can't be used.
On to the second question. We have already proven that the above formulae show that there is a limit as to the values of $b^2$ and $c^2$, and if there are infinite possible values of $b^2$ and $c^2$ then some of the must be non-integer. So by proof of contradiction for any real integer $k$, there can only be a finite amount of Pythagorean triples.
answered Nov 17 at 2:14
Kyky
418213
418213
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We have $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2.$
(1). The sequence $(2^2-1^2, 3^2-2^2, 4^2-3^2,...)$ of differences of successive squares is the sequence $(3,5,7,...)$ of odd numbers $>1$. So if $m $ is odd and $mgeq 3$ then we can find $b$ such that $(b+1)^2-b^2=m.$ That is, $2b+1=m.$ So let $b=(m-1)/2$ and $a=b+1.$ Then we have $$(a^2+b^2)^2=(a^2-b^2)+(2ab)^2=m^2+(2ab)^2.$$This does not work for $m=1$ as the member $2ab$ of the triplet would be $0.$
(2). If $m$ is even and $mgeq 4$, let $a=m/2$ and $b=1.$ Then we have $$(a^2+b^2)^2 =(a^2-b^2)^2+(2ab)^2=(a^2-b^2)+m^2.$$This does not work for $m=2$ as the member $a^2-b^2$ of the triplet would be $0$.
(3). If $x,y,z$ are positive integers with $x^2+y^2=z^2$ then
(3-i). It is easy to confirm that $1ne zne 2.$
(3-ii). We have $y<z$ so $ yleq z-1$ so $x^2=z^2-y^2geq z^2-(z-1)^2=2z-1geq 5$, so $x>2.$ Interchanging $x,y$ in this, we also get $y^2geq 5$ so $y>2$.
BTW. The general formula for ALL Pyth. triplets is ${k(a^2-b^2),,2kab,, k(a^2+b^2)}$ for $a,b,kin Bbb N$ with $a>b$.
add a comment |
up vote
0
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We have $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2.$
(1). The sequence $(2^2-1^2, 3^2-2^2, 4^2-3^2,...)$ of differences of successive squares is the sequence $(3,5,7,...)$ of odd numbers $>1$. So if $m $ is odd and $mgeq 3$ then we can find $b$ such that $(b+1)^2-b^2=m.$ That is, $2b+1=m.$ So let $b=(m-1)/2$ and $a=b+1.$ Then we have $$(a^2+b^2)^2=(a^2-b^2)+(2ab)^2=m^2+(2ab)^2.$$This does not work for $m=1$ as the member $2ab$ of the triplet would be $0.$
(2). If $m$ is even and $mgeq 4$, let $a=m/2$ and $b=1.$ Then we have $$(a^2+b^2)^2 =(a^2-b^2)^2+(2ab)^2=(a^2-b^2)+m^2.$$This does not work for $m=2$ as the member $a^2-b^2$ of the triplet would be $0$.
(3). If $x,y,z$ are positive integers with $x^2+y^2=z^2$ then
(3-i). It is easy to confirm that $1ne zne 2.$
(3-ii). We have $y<z$ so $ yleq z-1$ so $x^2=z^2-y^2geq z^2-(z-1)^2=2z-1geq 5$, so $x>2.$ Interchanging $x,y$ in this, we also get $y^2geq 5$ so $y>2$.
BTW. The general formula for ALL Pyth. triplets is ${k(a^2-b^2),,2kab,, k(a^2+b^2)}$ for $a,b,kin Bbb N$ with $a>b$.
add a comment |
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We have $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2.$
(1). The sequence $(2^2-1^2, 3^2-2^2, 4^2-3^2,...)$ of differences of successive squares is the sequence $(3,5,7,...)$ of odd numbers $>1$. So if $m $ is odd and $mgeq 3$ then we can find $b$ such that $(b+1)^2-b^2=m.$ That is, $2b+1=m.$ So let $b=(m-1)/2$ and $a=b+1.$ Then we have $$(a^2+b^2)^2=(a^2-b^2)+(2ab)^2=m^2+(2ab)^2.$$This does not work for $m=1$ as the member $2ab$ of the triplet would be $0.$
(2). If $m$ is even and $mgeq 4$, let $a=m/2$ and $b=1.$ Then we have $$(a^2+b^2)^2 =(a^2-b^2)^2+(2ab)^2=(a^2-b^2)+m^2.$$This does not work for $m=2$ as the member $a^2-b^2$ of the triplet would be $0$.
(3). If $x,y,z$ are positive integers with $x^2+y^2=z^2$ then
(3-i). It is easy to confirm that $1ne zne 2.$
(3-ii). We have $y<z$ so $ yleq z-1$ so $x^2=z^2-y^2geq z^2-(z-1)^2=2z-1geq 5$, so $x>2.$ Interchanging $x,y$ in this, we also get $y^2geq 5$ so $y>2$.
BTW. The general formula for ALL Pyth. triplets is ${k(a^2-b^2),,2kab,, k(a^2+b^2)}$ for $a,b,kin Bbb N$ with $a>b$.
We have $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2.$
(1). The sequence $(2^2-1^2, 3^2-2^2, 4^2-3^2,...)$ of differences of successive squares is the sequence $(3,5,7,...)$ of odd numbers $>1$. So if $m $ is odd and $mgeq 3$ then we can find $b$ such that $(b+1)^2-b^2=m.$ That is, $2b+1=m.$ So let $b=(m-1)/2$ and $a=b+1.$ Then we have $$(a^2+b^2)^2=(a^2-b^2)+(2ab)^2=m^2+(2ab)^2.$$This does not work for $m=1$ as the member $2ab$ of the triplet would be $0.$
(2). If $m$ is even and $mgeq 4$, let $a=m/2$ and $b=1.$ Then we have $$(a^2+b^2)^2 =(a^2-b^2)^2+(2ab)^2=(a^2-b^2)+m^2.$$This does not work for $m=2$ as the member $a^2-b^2$ of the triplet would be $0$.
(3). If $x,y,z$ are positive integers with $x^2+y^2=z^2$ then
(3-i). It is easy to confirm that $1ne zne 2.$
(3-ii). We have $y<z$ so $ yleq z-1$ so $x^2=z^2-y^2geq z^2-(z-1)^2=2z-1geq 5$, so $x>2.$ Interchanging $x,y$ in this, we also get $y^2geq 5$ so $y>2$.
BTW. The general formula for ALL Pyth. triplets is ${k(a^2-b^2),,2kab,, k(a^2+b^2)}$ for $a,b,kin Bbb N$ with $a>b$.
answered Nov 17 at 13:38
DanielWainfleet
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Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you.
– Kyky
Nov 17 at 1:55
I have no idea......
– Nithish Kumar R
Nov 17 at 1:59
If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers.
– rafa11111
Nov 17 at 2:06