solution for equation For $a^2+b^2=c^2$ such that $a, b, c \in \mathbb{Z}$
Do we know whether the solution is finite or infinite for $a, b, c \in \mathbb{Z}$?
We know $a=3, b=4, c=5$ is one of the solutions.
 A: Assuming $m,n$ be any two positive integers such that $m < n$, we have:
$$a = n^2 - m^2,\;\; b = 2mn,\;\;c = n^2 + m^2$$
And then $a^2+b^2=c^2$. 
A: To comment on whether the solution is finite or infinite note that if $(a,b,c)$ satisfy $a^2 + b^2 = c^2$, then so does $(ka,kb,kc)$ where $k \in \mathbb{Z}$. Hence, either no integer solution exists or infinite integer solution exists. 
You have already observed that $(3,4,5)$ satisfies $3^2 + 4^2 = 5^2$. Hence, infinite solutions exist.
Babak Sorouh has given the parameterization, which generates almost all possible solutions $a,b,c \in \mathbb{Z}^+$ (without scaling).
A: You can use your reasoning and conclude that there are infinite right triangles that you could draw, but that is not a good proof.
It is already known that $3^2 + 4^2 = 5^2.$ 
You can use this statement itself to prove that there are infinitely many Pythagorean Triplets. 
Let's prove that there are infinitely many Pythagorean Triplets in the form $(3k)^2 + (4k)^2 = (5k)^2,\ \, k \in \mathbb{N}$. 
(NOTE that you can apply this to any Pythagorean Triplet; it's a property).
We have $(3k)^2 + (4k)^2 = (5k)^2$, or $9k^2 + 16k^2 = 25k^2$. Divide both sides by $k^2$ (it is not zero). We have $9 + 16 = 25 \iff 25 = 25$. We just have proved that the statement is true for any $k \in \mathbb{N}$ and there are infinite elements in $\mathbb{N}$.
REMARK The general proof is related.
