Prove that there exist infinitely many pythagorean integers $a²+b²=c²$ Prove  that there exist infinitely many Pythagorean integers $a²+b²=c²$
My key idea is to show that there exists infinitely many integers that can be the length of the sides of a right triangle, but I fail at it.
Other try is that $\sqrt{a^2+b^2}=c$ and so it is an equation of a circle, so I tried to show that for every natural number c, there exist a and b such that $\sqrt{a^2+b^2}=c$ is the equation of a circle but I failed again.
How to prove it?
Thank you.
 A: You could simply look at multiples of $a$, $b$ and $c$, e.g. $2a$, $2b$ and $2c$.
A: Other than the multiples of any given Pythagorean integers ($ka,kb,kc$ with $a,b,c,k\in\mathbb N$ and $a,b,c$ being Pythagorean integers), you can try: $(m^2+n^2)^2=(m^2-n^2)^2+(2mn)^2$ with any $m,n\in\mathbb Z$.
A: It's a classical problem. 
$$a = m^2 - n^2$$
$$b = 2mn$$
$$c = m^2+n^2$$
Are all the solution of this form? We can think about the question in this term: suppose you are searching for such a triple, given $b$, thus you are searching $a^2+b^2=c^2$ so that $(c-a)(c+a)=b^2$. Now, searching for the prime factorization of b we have all possible factors for $c+a$ and $c-a$. If $b$ is even you have to choose both even factors for $c+a$ and $c-a$ in order to can solve the resulting system in integer numbers, and you will obtain a solution of the upper given form. If $b$ is odd you are forced to choose odds both factors and the system have a solution, but $a$ is forced to be even and you can change the role beetwen a and b, coming back to the upper given form. For each factorization of $b=2mn$ you will have a triple of the upper given form. However if you are given a triple you can reduce it to a fundamental form by dividing for $GCD(a,b,c)$ and then, choosing the even term as $b$, you will obtain its factorizazion in term of $c-a$ and $c+a$. So that it need, for c, to be of the form $c=m^2+n^2$. 
The problem, now is: when we are given c is it a sum of square? This is too a classical problem entirily solved by Fermat. For c even we will have that $a,b$ are too even. So that you can consider the problem for c odds. If c is given it will have a sum of square representation iff all the prime odd factors of $c$ are of the form $4k+1$. 
A: Once you found an example, you can multiply each a,b,c with any integer and get another solution
A: there are infinite points in a cartesian cordinate system around the origin which can be located by cordinates $(c\cos \theta ,c\sin \theta)$ and is seperated from the origin by a distance $c$. $c\cos \theta=a$ and $c\sin \theta=b$
i hope you understood.
