Arithmetic progression $(a,b,c)$ with $a^2+b^2=c^2$ and $\gcd(a,b,c) = 1$ I was wondering whether or not primitive Pythagorean triples could form an arithmetic sequence with their terms. I know that $(3,4,5)$ satisfies this condition, but I was wondering whether or not there are other ones, and how many there could be. Thanks!
 A: We have $$a=m^2-n^2$$ $$b=2mn$$ $$c=m^2+n^2$$ hence the condition $$a+c=2b$$ gives $$2m^2=4mn$$ implying $\ m=2n\ $ The only case where $\ m\ $ and $\ n\ $ are coprime, is $\ m=2\ $ , $\ n=1\ $ , giving $\ (3/4/5)\ $ . Hence there is no other primitive solution as desired.
A: I believe $(3,4,5)$ is the only primitive triple that works. Obviously any multiple of this triple, like $(6,8,10)$ would also work.
If we let $a$ be the least element in the triple, and assume an arithmetic progression, then the triple looks like $(a,a+d,a+2d)$, and these must satisfy $a^2+(a+d)^2=(a+2d)^2.$
Expand both sides and do some addition & subtraction to get
$a^2-d^2=2ad+2d^2$
$(a+d)(a-d)=2d(a+d)$
Now it is fine to divide out the $(a+d)$ factor since $a$ and $d$ are both positive.
So we have $a-d=2d$, which is the same as $a=3d$
In the $(3,4,5)$ example, we have $a=3$ and $d=1$.
However, the $a=3d$ equation shows that $a$ must be a multiple of $3$, hence only triples of the form $(3k,4k,5k)$ will work (for any $k\in\Bbb{N})$.
A: A Pythagorean triple in arithmetic progression would be $(a-d, a, a+d)$, where
$$(a-d)^2+a^2=(a+d)^2.$$
$$a^2-2ad+d^2+a^2=a^2+2ad+d^2$$
$$a^2=4ad$$
Given $a>0$, we have $a=4d$.
Thus, the triple is $(3d,4d,5d)$, and only one of those is primitive.
A: Among primitives
$$x^2 +(x+1)^2=(x+2)^2\\\implies x^2 +(x+1)^2-(x+2)^2=0\\\implies x^2 - 2 x - 3 = 0$$
Using the quadratice equation, we find
$$x\in\{3,-1\}\implies \text{the set of triples}\quad S=\{(3,4,5),(-1,0,1)\}$$
There are no other primitive solutions and, for non-primitives there are only mulitples of these like
$$(6,8,10),(9,12,15),(12,16,20)\cdots\quad (-2,0,2),(-3,0,3),(-4,0,4)\cdots$$
