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.
 A: Hint
$$2n+1=(n+1)^2-n^2$$
Hint 2 The square of your odd number is odd...
A: 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:
 $$\bigg\{\left(\frac{k+1}{2}\right)^2-\left(\frac{k-1}{2}\right)^2,2\cdot\left(\frac{k+1}{2}\right)\cdot\left(\frac{k-1}{2}\right),\left(\frac{k+1}{2}\right)^2+\left(\frac{k-1}{2}\right)^2\bigg\}$$
$$=\bigg\{\frac{k^2+2k+1-k^2+2k-1}{4},\frac{k^2-1}{2},\frac{k^2+2k+1+k^2-2k+1}{4}\bigg\}$$
$$=\bigg\{k,\frac{k^2-1}{2},\frac{k^2+1}{2}\bigg\}$$

A: To find pythagorean triples with any odd number :
 **formula** : (x^2)+((x^2-1)/2)^2=((x^2+1)/2)^2
        x=odd number

ex: x=5 (i.e odd number)
        (5^2)+((5^2-1)/2)^2=((5^2+1)/2)^2 

        (5^2)+((25-1)/2)^2=((25+1)/2)^2

          (5^2)+((24)/2)^2=((26)/2)^2

              (5^2)+(12^2)=(13^2)

                    25+144=169

                       169=169

A: Here is another amusing way to reach the same conclusion. The sum of the first $n$ odd numbers is $n^2$, i.e. $\sum_{i=1}^{n} (2i-1) =n^2$. That this is true can be seen by the properties of arithmetic series, which equal the number of terms (in this case $n$) times the average value of the terms (in this case $\frac{(2\cdot 1-1)+(2\cdot n-1)}{2}=n$; $n\cdot n=n^2$). The sequence of odd numbers features every odd number, and each odd number is therefore the difference between two successive squares. Of course, odd numbers can be the difference between squares that are not sequential as well, such as $8^2+15^2=17^2$.
