The discussion focuses on finding all natural number values of \(x\) such that the quadratic function \(f(x) = x^2 - 19x + 99\) results in a perfect square. The scope includes mathematical reasoning and exploration of conditions under which the function yields perfect squares.
The discussion does not present any clear consensus or resolution, as it primarily consists of a request for information and subsequent acknowledgments without further elaboration on the mathematical problem.
The discussion lacks detailed exploration of the mathematical steps or assumptions required to solve for the perfect squares, leaving potential gaps in reasoning.
Why?RLBrown said:This is equivalent to integer results for
f(k)=(19+Sqrt[4k^2-35])/2 where k is a Natural number.
eg. f(3)=10 and f(9)=18
Nicely done. (Bow)kaliprasad said:let
$y^2 = x^2 - 19 x + 99$
$\Rightarrow\, 4y^2 = 4x^2 - 76 x +396$ to make perfect square and avoid fraction
$\Rightarrow\, 4y^2 = (2x- 19)^2 - 19^2 + 396$
$\Rightarrow\, (2y)^2 = (2x- 19)^2 + 35$
$\Rightarrow\, (2x- 19)^2 -(2y)^2 = - 35$
$\Rightarrow\, (2x- 19+2y)(2x- 19-2y) = - 35$
without loss of generality we can assume y > 0
then we get factoring - 35
$(2x-19+2y, 2x - 19- 2y)$ =$ (35,-1)$ or$ ( 7, -5)$ or $(5,-7)$ or $(1, -35)$
or x = 18 or 10 or 9 or 1