# 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 ?

$$2(5^2)=1^2+7^2=(4-3)^2+(4+3)^2 \rightarrow (3,4,5)$$ $$2(13^2)=7^2+17^2=(12-5)^2+(12+5)^2 \rightarrow (5,12,13)$$ $$2(17^2)=7^2+23^2=(15-8)^2+(15+8)^2 \rightarrow (8,15,17)$$ $$2(25^2)=17^2+31^2=(24-7)^2+(24+7)^2 \rightarrow (7,24,25)$$

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.

$$m^2+n^2=2(p)^2$$ satisfying triplets make automedian triangles.
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: $$(x,y,z)=(28k^2-4k-1,20k^2-2k,4k^2+8k-1).$$