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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 ?

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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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We have $$x^2+x+z^2+z=2(y^2+y)$$ or $$(x+z+1)^2+(x-z)^2=(2y+1)^2$$ We got Pythagorean triples.

Let $x+z+1=m^2-n^2$ and $x-z=2mn$, where $m>n$ and $m$ and $n$ have a different parity.

Thus, $$(x,y,z)=\left(\frac{m^2-n^2+2mn-1}{2},\frac{m^2+n^2-1}{2},\frac{m^2-n^2-2mn-1}{2}\right).$$ Also, we can always take $\frac{m^2-n^2-2mn-1}{2}>0.$

For example, for $m=6k$ and $n=2k-1$, where $k$ is a positive integer, we obtain: $$(x,y,z)=(28k^2-4k-1,20k^2-2k,4k^2+8k-1).$$

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