<s>[INST] There are triads of perfect squares that are consecutive terms of arithmetic progression? Prove that there exist infinitely many triples of positive integers $ x , y , z $ for which the numbers $ x(x+1) , y(y+1) , z(z+1) $ form an increasing arithmetic progression.
$ \bigg( $ It is equivalent to find all triples of $ 4x(x+1)+1=(2x+1)^{2} , 4y(y+1)+1=(2y+1)^{2} , 4z(z+1)+1=(2z+1)^{2} $ $ \bigg) $
Note : I know $ \big( 1^{2} , 5^{2} , 7^{2} \big) $ , $ \big( 7^{2} , 13^{2} , 17^{2} \big) $ , $ \big( 7^{2} , 17^{2} , 23^{2} \big) $ , $ \big( 17^{2} , 25^{2} , 31^{2} \big) $ , but how i can found all triples ?
 [/INST] Looking at your examples,
$$ 2(5^2)=1^2+7^2=(4-3)^2+(4+3)^2  \rightarrow  (3,4,5) $$
$$ 2(13^2)=7^2+17^2=(12-5)^2+(12+5)^2  \rightarrow  (5,12,13) $$
$$ 2(17^2)=7^2+23^2=(15-8)^2+(15+8)^2  \rightarrow  (8,15,17) $$
$$ 2(25^2)=17^2+31^2=(24-7)^2+(24+7)^2  \rightarrow  (7,24,25) $$
Use any Pythagorean triplet such that $a^2+b^2=c^2$, then the identity at work is :
$$ (a-b)^2+(a+b)^2=2(a^2+b^2)=2(c^2) $$
So start with the general form of Pythagorean triplet and you may arrive at your desired construction.

Addendum :
$m^2+n^2=2(p)^2$ satisfying triplets make automedian triangles.
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