Inspired by Krishna Sharma

Too dificult. Just set x=a+b, a integer, b decimal between 0 and 1.

From ${{(a+b)}^{2}}-a(a+b)=1$ , we get to: $a=\frac{1-{{b}^{2}}}{b}$

Varying b from 0 to 1, we get many infinite solutions!

The point of this question, is that even though the equation "looks" quadratic because it only has terms of degree 2, there are actually infinitely many solutions.

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Solution: For each integer $n \geq 3$ , we can verify that $x = \frac{ n + \sqrt{ n^2 + 4 } } { 2 }$ is a solution. Hence, there are infinitely many solutions.

Are these all the solutions? Or are there others that I did not account for?

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Question that inspired this problem:

Find all positive solutions to $x^2 - x \lfloor x \rfloor - \lfloor x \rfloor ^{2} = 0$ .

It is easy to see $x \neq 0$ ; thus, we can rewrite the equation as $\lbrace x \rbrace = \frac{1}{x}$ , where $\lbrace x \rbrace$ is the fractional part of $x$ . Drawing the graph for both functions, we observe that $\frac{1}{x}$ intersects $\lbrace x \rbrace$ infinitely many times.

When x is an integer,

$x^2 - \lfloor x \rfloor x = 1$ becomes $x^2 - x^2 = 1$ which has no solutions.

Otherwise

$x^2 - x(x-q) = 1$ where 'q' is the frac(x).

$xq =1$ then has infinitely many solutions. Not sure if this is right.