<s>[INST] Prove that any odd number can be a leg in a Pythagorean Triple $(a,b,c)$ where $a^2+b^2=c^2$ is a Pythagorean triple.
My first thought was to do a proof by cases. I have three cases :


*

*a is odd, b is odd.

*a is odd, b is even.

*a is even, b is odd.


I'll just show case one :
$(2k+1)^2 + (2k)^2 = c^2$
I am kind of stuck relating how this answers the question. Thank you for any guidance in the right direction.
 [/INST] Hint:
Pythagorean Triples comes in the form: $(x^2-y^2,2xy,x^2+y^2)$ where $x$ and $y$ are any positive integers with $x>y$.
Can you use this to show that any odd number is possible?
Edit:
$(x^2-y^2,2xy,x^2+y^2)$ is equivalent to $(a,b,c)$ because:
$$\big(x^2-y^2\big)^2\ +\ \big(2xy\big)^2\ =\ \big(x^2+y^2\big)^2$$
Answer

 A Pythagorean Triple can be made from the values $(x^2-y^2,2xy,x^2+y^2)$
 We require one of the numbers to be odd and be any possible odd number.
 We will show that $x^2-y^2$ can represent any odd number with suitable choices of $x$ and $y$.
 So let $2n-1$ be the target odd number.
 If $x=n$ and $y=n-1$ then $x^2-y^2=n^2-(n-1)^2=2n-1$
 So we can create the Pythagorean Triple.

To expand upon that, given an odd number $k$ we can construct a Pythagorean Triple as follows:

 We have $k=2n-1$ so $x=n=\frac{k+1}{2}$ and $y=n-1=\frac{k-1}{2}$.
 Hence the Pythagorean Triple is:
 $$\bigg\{\left(\frac{k+1}{2}\right)^2-\left(\frac{k-1}{2}\right)^2,2\cdot\left(\frac{k+1}{2}\right)\cdot\left(\frac{k-1}{2}\right),\left(\frac{k+1}{2}\right)^2+\left(\frac{k-1}{2}\right)^2\bigg\}$$
$$=\bigg\{\frac{k^2+2k+1-k^2+2k-1}{4},\frac{k^2-1}{2},\frac{k^2+2k+1+k^2-2k+1}{4}\bigg\}$$
$$=\bigg\{k,\frac{k^2-1}{2},\frac{k^2+1}{2}\bigg\}$$

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