Find the smallest solution in positive integers of $x^{2} - 7 y^{2} - 9 = 0$ .

x = 4, y = 1

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x = 4, y = 2

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x = 5, y = 1

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None of these

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Solution

The correct option is A $x = 4, y = 1$
Given $x^{2} - 7 y^{2} - 9 = 0$
$\Rightarrow x^{2} = 7 y^{2} + 9$
The smallest possible positive integer for $y$ is $1$
For $y = 1$ , the value of $x$ is $4$
Therefore the smallest solution in positive integers is $x = 4, y = 1$
Therefore option $A$ is correct

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P is represented as $P = x_{a} y^{a} + x_{a - 1} y^{a - 1} + . . . . . x_{1} y + x_{0}$ where $0 < x_{0} . . x_{a} \leq y - 1$ . If x and y are positive integers and y=4, then which of the following could be a representation of an integer with y as a base?

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Find the smallest solution in positive integers of x2−7y2−9=0.

The correct option is A x=4,y=1Given x2−7y2−9=0⇒x2=7y2+9The smallest possible positive integer for y is 1For y=1 , the value of x is 4Therefore the smallest solution in positive integers is x=4,y=1Therefore option A is correct

Find the smallest solution in positive integers of $x^{2} - 7 y^{2} - 9 = 0$ .

The correct option is A $x = 4, y = 1$
Given $x^{2} - 7 y^{2} - 9 = 0$
$\Rightarrow x^{2} = 7 y^{2} + 9$
The smallest possible positive integer for $y$ is $1$
For $y = 1$ , the value of $x$ is $4$
Therefore the smallest solution in positive integers is $x = 4, y = 1$
Therefore option $A$ is correct