1. ## Little help on these if possible please?

I have some Pythagorean Triples proofs that I kind of have a hard time getting started on. I am just looking for a little insight to help me get started. Any help at all would be greatly appreciated. Here goes.

A pythagorean triple is a triple of positive integers (a,b,c) such that a^2 +b^2=c^2. In othere words, a set of three numbers is a pythagorean tiple if there exists a right triangle having these three thtegers as its side lengths. The triples one most commonly encounters when one first studies right triangles are (3,4,5) and (5,12,13).

(a) Are there othere pythagorean triples besides (3,4,5) which consist of three consecutive positive integers? Either give an exam;le of another such triple, or else prove n o other such triple can exist.

(b) Prove that there are infinitely many pythagorean triples (a,b,c) for which c -b=1. (Suggestion: direct proof by constructing them explicitly. Start by finding some small ones and look for a pattern. )

(c) Prove that there does not exist a pythagorean triple (a,b,c) such that a is odd, b is odd, and c sis even. ( Suggestion: Assume (a,b,c) is such a triple and derive a contradiction.)

On (a) it is pretty obvious that there is not any more triples that consist of three consecutive positive integers as a^2 + b^2 gets bigger much quicker than c^2. I think that this is proof by contradiction but not completely sure. Like I said any help would be greatly apprecitated.

I have some Pythagorean Triples proofs that I kind of have a hard time getting started on. I am just looking for a little insight to help me get started. Any help at all would be greatly appreciated. Here goes.

A pythagorean triple is a triple of positive integers (a,b,c) such that a^2 +b^2=c^2. In othere words, a set of three numbers is a pythagorean tiple if there exists a right triangle having these three thtegers as its side lengths. The triples one most commonly encounters when one first studies right triangles are (3,4,5) and (5,12,13).

(a) Are there othere pythagorean triples besides (3,4,5) which consist of three consecutive positive integers? Either give an exam;le of another such triple, or else prove n o other such triple can exist.
Let n be the smallest of the triple. then (n + 1) is the next in the triple and (n + 2) is the last. if there are consecutive positive Pythagorean triples, then they must satisfy:

$n^2 + (n + 1)^2 = (n + 2)^2$

solving this, we find that $n = 3$ or $n = -1$

$n = -1$ makes no sense, so only $n = 3$ works. thus the only Pythagorean triple that fulfills the conditions is 3,4,5

QED

(c) Prove that there does not exist a pythagorean triple (a,b,c) such that a is odd, b is odd, and c sis even. ( Suggestion: Assume (a,b,c) is such a triple and derive a contradiction.)
Assume, for the sake of contradiction, that such a triple exists.

Since $a$ is odd, $b$ is odd and $c$ is even, we can write: $a = 2n + 1$, $b = 2m + 1$ and $c = 2k$ for $m,n,k \in \mathbb {Z}$. Since they are a Pythagorean triple, we have that:

$(2n + 1)^2 + (2m + 1)^2 = (2k)^2$

$\Rightarrow 4n^2 + 4m^2 + 4n + 4m + 2 = 2 \left( 2 k^2\right)$

$\Rightarrow 2 \left( 2n^2 + 2m^2 + 2n + 2m + 1 \right) = 2 \left( 2k^2 \right)$

$\Rightarrow 2 \left[ 2 \left( n^2 + m^2 + n + m \right) + 1 \right] = 2 \left( 2k^2 \right)$

Thus we must have that: $2 \left( n^2 + m^2 + n + m \right) + 1 = 2k^2$

But $2 \left( n^2 + m^2 + n + m \right) + 1$ is odd, and $2k^2$ is even, so that cannot be. Thus we arrive at a contradiction.

Therefore, no such Pythagorean triple exists

QED

(b) Prove that there are infinitely many pythagorean triples (a,b,c) for which c -b=1. (Suggestion: direct proof by constructing them explicitly. Start by finding some small ones and look for a pattern. )
here are the first few. apparently i'm too tired to spot any helpful pattern, maybe you'll have more luck with it

(3,4,5), (7,24,25), (9, 40, 41), (11, 60, 61), (13, 84, 85), ...

note that all the first numbers are odd numbers greater than 1. you have a triple for all such odd numbers.

also, the numbers have this relationship.

a is odd, b is even and c is odd. if $a = 2n + 1$, $b = 2m \implies c = 2m + 1$ for $m,n \in \mathbb {Z}$, then:

$(2n + 1)^2 + (2m)^2 = (2m + 1)^2$ ................(1)

i guess you can search for a way to relate n and m. the problem will probably be solved if you accomplish that

EDIT: well, $n = \frac {\sqrt{4m + 1}}2 - \frac 12$ ................(2)

...does that help. i'm just feeling in the dark here, hoping to stumble on something

EDIT 2: well, i think i have at least the outline. first we prove that the triples have to obey equation (1). then we prove that for special integers m greater than 2, we can find an integer n that satisfies equation (2), so we have to find all integers m that cause n to be an integer. this of course happens when 1 + 4m is an odd perfect square

EDIT 3: ...i don't think this is going anywhere. i'm tired, wait for someone else, i got two out of three, that's good enough

5. Thank you Jhevon. You were a lot of help on these.

I have some Pythagorean Triples proofs that I kind of have a hard time getting started on. I am just looking for a little insight to help me get started. Any help at all would be greatly appreciated. Here goes....

(b) Prove that there are infinitely many pythagorean triples (a,b,c) for which c -b=1. (Suggestion: direct proof by constructing them explicitly. Start by finding some small ones and look for a pattern. )
...
Hi,

there are infintely many Pythagorean triples:

a) if (a, b, c) is a Pythegorean triple then $(ta, tb, tc), t\ \in \ \mathbb{N}$ is a Pythagorean triple too;

b) let $\boxed{\begin{array}{l}a=u^2-v^2 \\b=2uv\\c=u^2+v^2\end{array}}$ and $u>v~\wedge~u-v\text{ is odd}$

then you have a Pythagorean triple.
Examples
Code:
u    v    |   (a, b, c)
-------------------------
2    1    |   (3, 4, 5)
3    2    |   (5, 12, 13)
4    1    |   (15, 8, 17)
..   ...  |   ........

7. Originally Posted by earboth
Hi,

there are infintely many Pythagorean triples:

a) if (a, b, c) is a Pythegorean triple then $(ta, tb, tc), t\ \in \ \mathbb{N}$ is a Pythagorean triple too;

b) let $\boxed{\begin{array}{l}a=u^2-v^2 \\b=2uv\\c=u^2+v^2\end{array}}$ and $u>v~\wedge~u-v\text{ is odd}$

then you have a Pythagorean triple.
Examples
Code:
u    v    |   (a, b, c)
-------------------------
2    1    |   (3, 4, 5)
3    2    |   (5, 12, 13)
4    1    |   (15, 8, 17)
..   ...  |   ........
recall that c - b = 1. your last example does not follow that

8. Originally Posted by Jhevon
recall that c - b = 1. your last example does not follow that
Hi,

of course you are right. My post was meant to give the general pattern of constructing Pythagorean triples.

In this case you have to take:

$c-b=u^2-2uv+v^2=(u-v)^2=1$ that means u = v+1.

Thus all pairs (u, v) with this condition will give a Pythagorean triple like;

u = 4, v = 3 ==> (7, 24, 25)
u = 5, v = 4 ==> (9, 40, 41)

and so on...

9. Originally Posted by earboth
Hi,

of course you are right. My post was meant to give the general pattern of constructing Pythagorean triples.

In this case you have to take:

$c-b=u^2-2uv+v^2=(u-v)^2=1$ that means u = v+1.

Thus all pairs (u, v) with this condition will give a Pythagorean triple like;

u = 4, v = 3 ==> (7, 24, 25)
u = 5, v = 4 ==> (9, 40, 41)

and so on...
oh ok

10. It is shown in a standard number theory course that what Earboth posted in (b) is the complete list of Pythagoren triples.

We can use a bit of algebra on #3 ... rather than making a long list.

(b) Prove that there are infinitely many PTs $(a,b,c)$ for which $c -b\:=\:1$
Let $b \,=\,m$
Then $c \,=\,m+1$

The Pythagorean Triple would be: . $a^2 + m^2 \:=\:(m+1)^2$

. . which simplifies to: . $a^2\:=\:2m+1\quad\Rightarrow\quad a \:=\:\sqrt{2m+1}$

We want $2m$ to be one less than a square.
. . Then: . $m \:=\:4,\,12,\,24,\,40,\,\cdots$

We find that $m$ is of the form: . $2n(n+1)$ for any natural number $n.$

Therefore, there is an infinite number of such PTs.

(c) Prove that there does not exist a PT (a,b,c) such that
$a$ is odd, $b$ is odd, and $c$ is even.
Let $a \,= \,2p + 1,\;b \,= \,2q + 1,\;c \,= \,2r$ for natural numbers $p,\,q,\,r$
. . Assume that such a PT exists.

Then: . $(2p+1)^2 + (2q+1)^2\:=\:(2r)^2\quad\Rightarrow\quad 4p^2 + 4p + 1 + 4q^2 + 4q + 1 \:=\:4r^2$

. . $\Rightarrow\quad4(p^2+p+q^2+q) + 2 \:=\:4r^2\quad\Rightarrow\quad 4r^2 - 4(p^2+p+q^2+q) \:=\:2$

. . $\Rightarrow\quad4(r^2 - p^2-p-q^2-q) \:=\:2$

The left side is a multiple of 4; the right side is not.

We have reached a contradiction . . . Q.E.D.