Question

Given that x and y are positive integers, solve the equation x² - 4y² = 13​

Answer 1

Answer:  x = 7 and y = 3

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Explanation:

Apply the difference of squares rule

x² - 4y² = 13

x² - (2y)² = 13

(x - 2y)(x + 2y) = 13

Since x and y are positive integers, this means x-2y and x+2y are both integers as well.

The value 13 is prime. Its only factors are 1 and 13

Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:

• A) x-2y = 1 and x+2y = 13
• B) x-2y = 13 and x+2y = 1

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Let's consider case A

We have this system of equations

$\begin{cases}x-2y = 1\\x+2y = 13\end{cases}$

Add the equations straight down

• x+x becomes 2x
• -2y+2y becomes 0y = 0 which goes away
• 1+13 becomes 14

Therefore we have 2x = 14 solve to x = 7

From here, plug this into either equation to solve for y

x-2y = 1

7 - 2y = 1

-2y = 1-7

-2y = -6

y = -6/(-2)

y = 3

You should get the same result if you used x+2y = 13

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Since we've found that x = 7 and y = 3, notice how case B is not possible

Example:  x-2y = 13 becomes 7-2(3) = 13 which is false.

Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.

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Let's check those x and y values in the original equation

x² - 4y² = 13

7² - 4*(3)² = 13

49 - 4(9) = 13

49 - 36 = 13

13 = 13

The answer is confirmed.