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anemone
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Find all values of $x$ such that $f(x)=x^2-19x+99$ is a perfect square for all $x\in N$.
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
The formula for finding the perfect square of $f(x)$ is $(x-\frac{b}{2})^2$, where $b$ is the coefficient of the $x$ term.
The perfect square of $f(x)$ exists for a given natural number $x$ if the discriminant of the quadratic equation $x^2-19x+99=0$ is a perfect square. The discriminant is given by $b^2-4ac$, where $a$ and $b$ are the coefficients of the quadratic equation.
The steps for solving for the perfect square of $f(x)$ for a given natural number $x$ are as follows:
No, the perfect square of $f(x)$ cannot be negative. It will always be a positive number or zero.
No, the perfect square of $f(x)$ is defined for all natural numbers. However, the value of $x$ may vary depending on the given quadratic equation.