Solving for Perfect Squares of $f(x)=x^2-19x+99$ for All Natural Numbers $x$

[anemone]

Find all values of $x$ such that $f(x)=x^2-19x+99$ is a perfect square for all $x\in N$.

 

[RLBrown]

Spoiler

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

 

[topsquark]

Spoiler

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

Why?

-Dan

Oops. Missed the obvious. Thanks.

-Dan

 

[kaliprasad]

Spoiler

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

 

[topsquark]

Spoiler

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

Nicely done. (Bow)

-Dan