question
Show that there exist infinitely many non similar triangles such that the side-lengths are positive integers and the areas of squares constructed on their sides are in arithmetic progression.
Solution
let the sides be $a,b,c$ , let $a \ge b\ge c$ so , we have $a^2+c^2=2b^2$. put $a/b=x ,c/b=y, x,y>0$ so , we get $x^2+y^2=2$....() , with $x,y\in Q, x,y>0$ now , consider a particular solution $(p,q)$. consider the slope (k) of the line joining $(p,q)$ & $(x,y)$. $k=\frac{y-q}{x-p}$ , now express $y$ in terms of $k,p,q$ and put that in (). then appying Vieta's relation , we can easily solve the problem.we must keep in mind the problem conditions for choosing $k$ carefully. a such set of solution is given by $(a,b,c)=(k^2-1+2k,k^2+1,k^2-2k-1)$. ---------(###) we must choose $k\in N , [k\ge5] $ carefully such that , they are consistent with the problem condition. it is quite obvious that , we can make infinitely many non-similar triangles satisfying the condition from this family with suitable values of $k$ [the proof of the last sentence is trivial ,so i am not posting it. it uses basic divisibility and triangle inequality
Source
https://artofproblemsolving.com/community/c6h464522p2602940
Doubt
Please some body explain how vieta relation apply here and how the equation -----------(###) comes And why $k\in N , [k\ge5] $.
The given solution hints at a general method to solve an equation like $$x^2+y^2=2 \tag{1}$$ for rational $x$ and $y$, but the explanation is not so clear.
The trick is that of choosing a particular solution $(p,q)$, for instance $(1,-1)$, and parameterise all other solutions $(x,y)$ by the slope $k$ of the line through $(1,-1)$ and $(x,y)$, which is a rational number. The equation of that line is $$ y+1=k(x-1), \quad\text{that is}\quad y=kx-k-1. $$ Substituting this into $(1)$ gives $$ (1+k^2)x^2-2k(k+1)x+k^2+2k-1=0 $$ and factoring out the solution $x=1$ (or using Vieta's rule for the product of roots) we are left with $$ x={k^2+2k-1\over1+k^2} \quad\text{and}\quad y={k^2-2k-1\over1+k^2}. $$ For the purpose of the given exercise you can simply take integer values of $k$, with $k\ge3$ (not $5$) to get positive solutions. But for a more general solution you may want to substitute $k=m/n$ and express $a$, $b$, $c$ as a function of integers $m$ and $n$.