Question

not a question
Art of Problem Solving
Methods of Solving Diophantine Equations

2. The Parametric Method

Answer 1

In many situations the integral solutions to a Diophantine equation

\(\color{green}{ f(x_{1}, x_{2}, . . . , x_{n}) = 0 }\)

can be represented in a parametric form as follows:

\(\color{green}{ x_{1} = g_{1} (k_{1}, . . . , k_{m}) \ , x_{2} = g_{2} (k_{1} \ ,. . . , \ k_{m}) \ , \ \ . . . \ \ , \ x_{n} = g_{n} (k_{1}, . . . , k_{m}) }\)

where \(\color{green}{ g_{1}, g_{2} \ , \ . . . \ , \ g_{n} }\) are integer-valued m-variable functions and \(\color{green}{ k_{1}, . . . , k_{m} ∈ Z }\).

.....The set of solutions to some Diophantine equations might have multiple parametric representations. For most Diophantine equations it is not possible to find all solutions
explicitly .In many such cases the parametric method provides a proof of the existence
of infinitely many solutions.
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Here is a Problem (by myself) to Introducing this method:

Prove that there are infinitely many triples \(\color{red}{(x, y, z) }\) of positive integers such that \(\color{red}{ x^2 + y^2 = z^2}\) .

Well ; for this case \(\color{red}{ x^2=z^2-y^2=(z-y)(z+y) }\) Setting \(\color{red}{ z-y=1}\) gives \(\color{red}{ z=y+1}\) and the equation becomes \(\color{red}{ x^2=2y+1 }\) .

Note that \(\color{red}{2y+1 }\) is an odd number so \(\color{red}{x}\) must be an odd number too ; let \(\color{red}{ x=2k+1}\) it gives \(\color{red}{(2k+1)^2=2y+1 }\) or \(\color{red}{ y=2k(k+1) }\) and we will get \(\color{red}{ z=y+1=2k^2+2k+1 }\) .

We obtain the infinite family of solutions : \(\color{red}{(x, y, z)=(2k+1 \ , \ 2k^2+2k\ , \ 2k^2+2k+1) \ \ k=1,2,3,… }\)
are answers for Pythagorean equation.